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"Time for that on the way to the station." air max plus 2015 air max plus 2015 "Now do you repay a hundredfold all the suffering of these years," he said. "Living or dying, I am content."
"It was old Parson Pralph walking back from Hereford to his Vicarage at Tarrington." "It is nothing but trouble, Signore. Such a fine young man, and the piccola loved him so! It will surely place the Maestro among the saints. Four masses for his soul, Signore; and those priests are such thieves. I said 'No lesson,' but the Maestro is a mule for having his own way. Let him teach, say I; it will divert his mind! There, Signore, go in with you! But I always thought it would come; four times I heard the cock crowing, a bad sign, as Saint Peter knew. There, there! the Madonna aid us!" "No, only a legal document just issued by Prince Rupert. I saw him not long since at Hereford." air max plus 2015 "Hal's gone out. I suppose you've been——" "Just in time for afternoon tea," said the Major gallantly, and threw open the door. "Please to enter a bachelor's den, Miss Hedge." "We owe Lieutenant Harford a debt of gratitude, my lord," said the Rector, "not only for rescuing Helena from the vile scheme of Colonel Norton, an ill-conditioned neighbour of her father's, but for all that he did for my dead kinsman. Helena hopes shortly to be in London under the charge of her godmother, Madam Harford, a kinswoman of the lieutenant's, and she will do her utmost to obtain his exchange. But we trouble your lordship too much with our affairs. In what can I serve you, my lord?" air max plus 2015 But did mobility, backed by the rifle, inspire dash in the Boers? At this period, except in isolated cases, no. The inertia, so disproportionate to their tactical flexibility and brilliant skirmishing skill, was never more apparent 72than at the close of this battle outside Ladysmith, when White began his retreat. No such opportunity was ever to present itself again for a really decisive victory. Joubert regarded the action as a defensive action, and had issued general orders against a pursuit. On the other hand, a simple burgher, Christian de Wet, had inspired the one genuinely aggressive enterprise which distinguished the Boer movements on this day—namely, the attack and capture of the detached force at Nicholson's Nek. This—like the capture of Majuba nearly twenty years earlier—was a feat of stalking pure and simple, with which the horse had little to do, save that it bore the riflemen rapidly from a distant part of the field into the outermost fringe of the zone of combat. The outermost fringe—that is the point to watch. Could horses penetrate the inner fringe under rifle-fire and so precipitate the decisive phase of a conflict? While waiting for the answer let us be sensible and remember that after all the main thing is to win fights. Galloping under fire is only a means to an end. Stalking under fire requires nearly as much dash to be effective. The day was hot and drowsy, and Beatrice, clothed in black--for she paid her stepfather the compliment of wearing mourning--sat on an old stone seat, between two yew trees cut in the shape of peacocks. Before her, on a slight rise, rose the mellow brick walls of the Grange, covered with ivy. A terrace ran along the front of the house, and over the door was the mouldering escutcheon of the Paslow family. What with the queer pointed roofs, the twisted stacks of chimneys, the diamond-paned casements, and the prim gardens, the place looked particularly delightful. A poet could have dreamed away his days in this rustic paradise, and Beatrice felt as though she were in the land of the Lotos-eaters. But even as she slipped into vague dreams, she pulled herself up, and shunned the enchanted ground. There was sterner work to do than dreaming. Before she could become the mistress of this castle of indolence, and wife of its master, it was necessary to lift the cloud which rested on the place. To do so, she would have to begin by questioning Mrs. Lilly, and impatiently awaited the arrival of that worthy soul. "We can. But, if we do, I'll have more time to make a fool of myself, and I've done that pretty thoroughly." As they passed he turned his gleaming eyes full upon them, and though she gave him a cheerful "Good morning," he made no reply, only touching his hat in a grudging and reluctant fashion. "I won't trouble you. Thanks all the same." "Oh! Lord Harry Dalblane knows naught of humanity," said Captain Tarverfield, laughing.
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"Taking a look at Tom? Be careful; you'll make him jealous." She slipped her small hand into his. "I can't tell you how good it is to see you." CHAPTER XIV. THE NEW LAZARUS
With her face averted she said: "At first you thought you had only to love and she'd love you in return—wasn't that it? With you to love her has been a mission; that's where you're different from other men. Other men start by flirting—they intend the run-away right up to the last minute; then they find themselves caught But you—— It takes an older woman than Desire to understand. You're so impetuously in earnest, you almost frighten her. You're such a dreamer—the way you were about the marriage-box. You always take a woman at her word; and a woman, when she's loved, means most by the things she leaves unsaid. What happened to the marriage-box after you found me out?" And gibbets keep the lifted hand in awe; "Ay, my dear, but I come from Worcester with yet later news of defeat. My Lord Essex, who is in command of the rebel army, entered Worcester and has taken possession of the city. With my own eyes I saw his vile troops quartered in the cathedral; the knaves had no sort of reverence, and have stabled their horses in the cloisters. But there! I could not offend your ear by describing the scene. Would that I had the hanging of them! They should have but short shrift!" Norton veiled his annoyance by a laugh.
"Because"---- he swallowed something, and told what was evidently a lie, so glibly did it slip out. "Because I should be sorry to leave you to starve." "Gabriel, indeed! Rather than be protected by him I would throw myself on the mercy of any other man in England! Dr. Rogers says I did well and loyally in vowing never to see 'my old friend,' as he calls him, again, and if he dares to seek me out, I will make him suffer as he has not suffered yet." The Bishop at that moment caught sight of his manuscript, and, to Gabriel's disappointment, said no more about Hilary. "My commentary on the Colossians is complete," he remarked, turning over the leaves with a loving touch. "This afternoon I place it in the printer's hands." nike shoes for sale "Oh, we always call dinner 'luncheon,' and have it in the middle of the day; it saves trouble, and it is less worry." Then, after a moment's pause: "I wish we had a lobster, but I don't think there is one. I know there is a beefsteak." It was a sensible resolve, promptly made and admirably executed. At the moment when French formed it he was about a mile distant from the Modder and about two miles from the Nek. His division, in column of brigade masses, had been checked by the fire of the two Boer guns posted on the western or left-hand ridge, about 3,000 yards away, and by rifle-fire from the nearest part of the eastern ridge, about 1,500 yards away. All seven batteries of Horse Artillery, supported by two batteries of the Infantry division and two naval guns—fifty-six guns in all—had opened on the two ridges and the devoted pair of Boer guns, and had temporary silenced the latter. It was now that French ordered the charge, and while it lasted, all but two horse batteries, which were kept in reserve, continued to bombard the Boer positions. Gordon's brigade, less two squadrons, which were engaged on the flanks, led the way, deployed in extended order—eight yards between files, twenty yards between front and rear rank—pace, fourteen miles an hour. Broadwood's brigade came next, 800 yards behind, and the other two brigades one of Cavalry, the other of mounted riflemen followed, though exactly at what interval and in what formation we are not told precisely. But I think we may assume that the fully deployed charge was made only by Gordon's brigade, and that, at any rate, this was enough to secure the object in view.[24] nike shoes for sale He looked irresolutely at the phial in his hand, and then turned to Avenza, who Was still counting the feeble pulsation of the blood. "Oh, but it's not the delay I'm sorry for, it's the—it's the having to say that—I can't say what he wants me to say." nike shoes for sale The last word seemed to reach Gabriel. He opened his eyes for a moment and caught a misty glimpse of Dr. Coke and Hilary with a rough stone tower wall and a deeply splayed narrow window in the background. Was he once more a prisoner in St. George's Tower at Oxford? The horror of the thought roused him. Then he noticed that he was lying in a bed on the floor, and that they had removed his buff coat, a perception which vaguely troubled him. .. nike shoes for sale "I can't say. I think"--Vivian hesitated--"I think that he wishes to get possession of the Grange."
She cut him short. "Feeding-time. You ask too many questions." Teddy turned; his face was glowing with excitement. "It's Vashti." air max plus 2015 "He wasn't a beast. He was quite nice. You came near seeing him." nike shoes for sale nike shoes for sale "Not—not Bevan?" "The man's gone," said Chamberlin; "he departed Monday morning." She broke off and glanced over at him, making her mouth sad. "You do sit far away." When he made no motion to accept her invitation, she smiled the unreserved smile of friendship. "Look here, if I come half way over, will you?"
"She cheered us all with her sweet voice on Christmas night," he remarked as he rose from the table and led his guest into the library; "and better than all her other songs was a carol which she told me you had taught her as a child." He nodded. Try as he would, he could not disguise the fact that he was in love with her. A light shone in his eyes. This seemed no game, but a natural preliminary to something that must happen. She was indignant when the custodians of the apartments took it for granted that they were an engaged couple. She ungloved her hand that they might see for themselves that the ring was lacking. "It's for my mother," she explained. "Yes, I like the apartment; but I can't decide till my mother has seen it" She referred to Teddy pointedly as "My friend." The janitors looked knowing. They smiled sentimentally and put her conduct down to extreme bashfulness.
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There is little that need detain us in the further advance to Bloemfontein. It began on March 10, in three parallel columns under French, Tucker, and the Commander-in-Chief, and ended in the occupation of the capital on the 13th. Demoralization turned to genuine despair in most of the burghers who fled from Poplar Grove. Whole commandos melted to a shadow through desertion. It was only through the agency of reinforcements brought up by De la Rey, notably the Transvaal Police Zarps , that a show of resolute opposition could be organized by De Wet. At Abraham's Kraal or Driefontein , eighteen miles east, where, on March 10, the next stand was made, and where French commanded, the principal interest lies in the fine Infantry attack of the sixth division towards the evening, and the stubborn defence made by the small body of Zarps on the Driefontein Kopjes. An attempt by the Cavalry a little earlier in the day to turn the enemy's left was unsuccessful, and the final pursuit came to nothing. "Ah, yes!" The thin man ceased from searching through his pockets and heaved a sigh. "I lose everything. Suppose I'll go on losing till the grave shuts down on this body o' me—and then I'll lose that. My 'air began to come out before I was twenty—tonics weren't no good. Now I always 'ave to wear a 'at—do it even in the 'ouse, unless I'm reminded. And then, as you say, there was poor 'Enrietta. I'm always wondering whether I really lost 'er, or whether——" nike shoes offers online My dear, He was sorry the moment he had said it. A spark of his resentment caught fire in her eyes. He watched it flicker out. She smiled wearily, "So you think I haven't any passion!—Oh, well, we're going to have fine times, now that you've begun to criticize.—I'm sleepy. I think I'll go to bed." nike shoes offers online He could imagine the mischief that had crept into her gray eyes as she had listened to whatever Hal had told her. Why didn't she send for him? nike shoes offers online "How old are you? Sixty-three? stab , that's what you say, but you know very well you were born sixty-five years and six months ago. Wake up stab, stab , you must not go to sleep. Sixty-five—five years more and you will be seventy; fifteen years more and you'll be eighty, and you are in love stab, stab, stab . I'll teach you to eat sweet cakes and ice creams; I'll stab teach you to drink Burgundy. And you dared to call me Arthritic Rheumatism the other night, you stab dared! Now, go to sleep stab, stab ... wake up again, I want to speak to you," etc., etc., etc. nike shoes offers online
"Right follows might 324Another point: What are "masses"? I take the word from Bernhardi, who seems not to contemplate shock without great masses, the greater the better. Between the mass and the patrol, where is shock to come in? The patrol, where combat is necessary, according to "Cavalry Training," acts chiefly with fire, and Bernhardi says the same. For what size of unit does shock begin to be specially applicable? "Cavalry Training" is dumb. Bernhardi, more frank, as usual, seems to imply that it is really applicable only to very large masses. But why this mystery? Why should not even patrols use it? Shock is silent, and therefore suitable. Does it make any difference whether the unit is 10, 50, or 100 strong, or 500, or 1,000, or 5,000? From the arme blanche point of view it is wiser to leave the question unanswered. The answer would throw a flood of light on the "peculiar" conditions of South Africa, where during a great part of the war the numbers engaged were comparatively small. "Wide open." "Ah!" said Major Ruck, smiling more broadly than ever. Horace and Teddy drew up to the logs, toasting themselves and sitting near together. There was a distinct atmosphere of disappointment. They glanced at each other occasionally, saying nothing. It was an odd thing, Teddy reflected—the men whom he met at Vashti's apartment rarely had anything to say to each other. They met distrustfully as the women's friends. They never talked of their interests or displayed any curiosity; yet most of them were distinguished in their own line and would have been knowable, if met under other circumstances. As he dressed he molded the future to any shape he chose, rolled it into a ball and molded it afresh. Now that he had kissed her, all things were possible. His interest in all the world was quickened. His work and success again became important. He thought of her thin little high-heeled shoes, her dancing decorative way of walking, the costly frailty of her dress. He would need money—heaps of it—to marry her. [Pg 198] She glanced down at her green tweed suit. "Not dressy enough. Besides," she consulted the watch on her wrist, "it's nine.—Oh, I know; let's visit Fluffy. We'll catch her between the acts."
Nothing is big in the Southwest; transactions of millions are disposed of while one eats a flap-jack. In an hour the stranger had acquired The Flim Flam Murphy interest which was vested in Chicken Bill; in two hours that immoralist was speeding by vague trails to regions new, forty-five thousand dollars in his belt and a soreness in his heart. "A matter of opinion," she said. "Signor Hugo will attend to all that, Avenza." "Horace'll have to own now that she was wise in cultivating other friendships. Poor old Horace!—And all those bills will be covered up to-morrow with some new great success. Such is fame!—Fluffy's so discouraged." As she stooped to his knee to write, her hair fell back, exposing the whiteness of her neck. The familiarity with which she was filling these last moments sent all his dreams soaring. The daintiness, the slimness, the elfin beauty of her quickened his longing. His instinct told him that she was hoping that he would kiss her; but he guessed that, if he did, she would repulse him. "You go too fast for me," she had said. Once again his imagination wove a magic garment and flung it about her shoulders. There was no one like her. She was called Desire because she was desired. If love could compel love, she should come into his life. He vowed to himself that he would win her. nike shoes offers online With great tact, the hostess contrived in a few words to tell Mr. Geers the state of affairs, and the good-natured owner of Garnons undertook, in his cordial, friendly way, to talk matters over with the Vicar. "And did she?" nike shoes offers online She gazed out of the window. "It's all about us." Then, with a laughing glance at him, "I don't know whatever you'd do, if you didn't have me to write about." nike shoes offers online "It's in nearly the next street, Teddy." "I'm not the wife for you. I've given you far more misery than happiness."
Sir Ralph Hopton sighed as he refolded the letter; it had only made him crave more passionately for an end of the war which was dividing England. He glanced a second time at Gabriel, struck by something familiar in his face. "Are you not one of the Herefordshire Harfords?" he asked.
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"Oh Lord! I thought some one had shot a bullet into my foot. Give me the medicine,[Pg 283] quick, and send round for Carter. I must have some opium, or I won't sleep a wink." "I wouldn't dare to trust myself so much with any other man," she told him. "How do you know that this wire is an answer, Durban?" cheap nike air max 90 trainers Like tail follows kite." She caught the boy to her breast. She was trembling all over and on the verge of tears. When she had climbed into the wagonette, with Ruddy seated beside her, and had lumbered slowly out of the farmyard, she left Teddy wondering: Why had she said "a good woman"? As though there was any doubt that little Desire would grow up good! Hilary, in deep disgust, promptly drew the leathern curtain across the window, but though she could thus shut out the hideous leering face of the pseudo-patriot, she could not banish his words, which persistently rang in her ears as the coach lumbered out through Byster's Gate and along the rough road to Whitbourne; nor could she shut out the mental picture which the words conjured up, the picture of Gabriel Harford with a rope about his neck. cheap nike air max 90 trainers For some days past, ever since Verneede's fiasco in fact, Leavesley had been very much down in the mouth. cheap nike air max 90 trainers "Yes—'The Laurels,' John's Road. Have you any money?" He at once became repentant. "I only noticed her when I was lonely," he excused himself; "I thought she was like you." cheap nike air max 90 trainers
"Ah, yes." Teddy sighed sentimentally. His sigh said plainly, "Whatever else I don't know, I know that." He cushioned his face against the pillow. "But what I meant," he explained, "is supposing one hasn't any money, and one's father can't give one any, and one wants to be with some one every minute, and—and very badly. Would they live together then from the beginning?" "He sent most impertinent letters," said Fanny, "and I told father not to mind them, then James came." Leavesley glanced at the office clock, it pointed to quarter past ten, and Fanny was due at one. "Something for me?" she guessed. air max plus 2015 "Oh, d——n!" said Leavesley. "I'd love to," said Miss Lambert. "I haven't been up the river for ages; let's have a picnic." nike shoes offers online He found himself gazing into the humorous blue eyes which had discovered him playing tricks with his engine before the house in Regent's Park. nike shoes offers online Meanwhile in the tiled cottage to the east of the churchyard an old clergyman, in the last throes of a lingering and painful death, faintly gasped the words, "Lord, how long?" PART III CHAPTER I AN ASSIGNATION
He drove on, dreaming, through the summer night.
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"About Pallanza?" With the secret magic and passivity of night, the city had undergone a change. It had lost its haste. It went on tiptoe now. Tall buildings stood silent as tombs, quarried from the granite of the dusk. Streets had become orientalized. A spirit of poetry was abroad. Over the turrets of this Babylon of a day the wings of Time brooded, shadowing its modern glare with the pomp of a sombre and mysterious austerity. It had become a metropolis of dreamers, as fitting a stage as Florence for any tale that love might choose to tell. As he turned, the child ran forward and flung her arms about his neck. "Oh, Teddy, this person says perhaps she'll help us to find her." Then, in a whisper, bringing her face so dose that the thistledown of her hair brushed his forehead and his whole world sank into two gray eyes, "The Princess wasn't very nice this morning—not modest, so this person says. But you don't mind—say you don't I did so want to be like you and to do everything that boys do," and then, long drawn out, when he thought her apology was ended, "Teddy." nike trainers womens cheap "I'm cold," she whispered mournfully. "I'm trying to feel what I ought to be feeling, but I can't. I'm disappointed. God left something out when He made me. If only you weren't so fine, but—— My dear, you're better than any man I ever met. I couldn't be good the way you are, and I'm ashamed to be worse. Sometimes I'm almost bitter against you for your goodness. My beautiful mother.—I'm all she has. And there's your family. I haven't any. I've missed so much. Surely you under-stand?" There was much, too, to divert his mind when they reached Brampton Bryan Castle. He quickly fraternised with Ned, the eldest son, and quite lost his heart to sweet-faced Lady Brilliana, Sir Robert's wife. She had travelled much, and had lived for many years in Holland, so that her talk was infinitely more easy as well as more interesting than that of any other lady he had met. Moreover, while Sir Robert was of the somewhat hard and narrow Puritan school, she was surely the gentlest Puritan dame that ever breathed, and seemed full of kindness to all, whatever their views. He began on that day, as they wandered about the castle, a friendship with Ned which was to last all through his life, and his pity for his father was great when, on returning from the most delightful scramble about the battlements, they found Sir Robert Harley still discoursing on the vestments—and what he indignantly called the 'altar-ducking'—which Archbishop Laud had ordered in Hereford Cathedral. Beatrice raised her arms, and dropped them with a helpless air. She seemed to be more than ever surrounded by mysteries, and Durban, who was able to explain, insisted upon holding his tongue. At all events, her mind was set at rest regarding the honesty of Vivian; and she thought it best to take the old servant's advice, and possess her soul in patience until such time as he chose to tell her the truth, whatever that might be. But it was all very puzzling, and her head ached with the effort to think matters out. After a time Durban persuaded her to lie down, which she did very willingly, being quite prostrate after the terrors of the past few days. "I shan't catch cold either. And if I do come out you'll only be cross with me." nike trainers womens cheap He knotted his hands, trying to reduce his imagination to words. "It's the women who aren't like you, Dearie—the women who love themselves. They feed among lilies; the soul of love is in 'em, but they won't let it out They're gardens enclosed, fountains sealed, springs shut up. Now are you getting there? The symbolism of it caught me. There I have her, just as she is in her bang-up modern dress, feeding among the lilies of an Eastern garden. Everything's heavy with fragrance, beautiful and lonely; the hot sun's shining and nothing stirs. The windows of the harem are trellised and shut. From under clouds the north and south wind are staring and puffing their cheeks as though they'd burst. Through a locked gate in the garden you get a glimpse of an oriental street with the dust scurrying; but in my sister's garden the air hangs listless. The fountain is dry; the well is boarded over. And here's the last touch: halting in the street, peering in through the bars of the gate is the figure of Love. The woman doesn't see him, though he's whispering and beckoning. Love's got to be stark naked; that's how he always comes. Because he's naked he looks the same in all ages. D'you get the contrast between Love and the girl's modern dress? There's where I'll need you, Teddy." "That's hard lines on you." He sank his voice sympathetically. "You'll miss her awfully." "We will discuss the matter no more," he said, presently, "but to-morrow in church let us try to meet in all sincerity as fellow-worshippers. Now I will show you the piece of oak we spoke of, and you shall take the measurements for the corner cupboard." nike trainers womens cheap "You can if you really care." nike trainers womens cheap
"I'm a total stranger," she had said. "Go on with your breakfast, Mr. Gurney, I've had mine. I'll watch you. Well, I've heard, and so I've dropped in to see what I can do. You mustn't mind me; trying to be a mother to everyone's my foible. Now, first of all, you can't have that boy in the house—boys are nice, but a nuisance. They're noisy." Gabriel roused himself from the despair into which he had fallen, and picking his way cautiously across the forms of the sleeping prisoners, sat down beside the man with the fever. There was still a little water left in the earthenware mug near him, and, raising the poor fellow into an easier posture, he held this to his parched lips. nike trainers womens cheap She spoke again, slowly. "If ever Vashti were to accept you, it would be the worst day's work. The gods you worship are different. Hers are—hers are worthless." "Go away!" said his uncle. "But you don't—you wouldn't." "Did it go well?" asked Teddy. nike trainers womens cheap nike trainers womens cheap "Sir, indeed I dare not let you run such a risk; if aught should befall you what would become of Hilary?" said Gabriel, his eyes full of anxiety.
"What was his crime?" asked Hilary. "I'll bathe," he consented, "if you'll come with me. But aren't you hungry?" nike air max for sale "Yes. I'm here to see October put on; that's my new play in which Miss Audrey is acting. What are you doing?" Then, because Teddy hesitated, "Perhaps I oughtn't to ask." To pass the time, Beatrice swept out her bedroom and made the bed, then attended to the garden. Every now and then she would glance at the gate, expecting that Vivian Paslow would enter. But by twelve o'clock he had not come, and she felt very disappointed. Then she began to feel alarmed. What if he had met the man and had fought with him? What if the man had hurt him? She asked herself these questions, and half determined to go over to Convent Grange in order to get answers. But she did not wish to leave The Camp until Durban came back, since Alpenny was absent. Still the desire to hear and see Paslow was overwhelming, and she was just about to yield to her curiosity and leave The Camp to look after itself when she heard the rapid vibration of the electric bell, and knew that someone was at the gate. In a moment she was flying across the lawn, her heart beating and her colour rising. air max plus 2015 [Pg 279] "I am glad that Mr. Paslow arrived at the moment," said Durban, when the story was ended. "He and the Major now understand one another." "One moment, Marchese--I wish to speak with this Signor for a few minutes." nike air max for sale But to have flirted with a child! To have filled him with dreams and to have broken down his shyness! As she sat there with his box, labeled MARRIAGE, in her lap, she wondered what was best to be done. If she told him it was a jest, she would rub the dust off the moth-wings of his faith forever. There was only one thing: to continue the extravagant pretense. The Captain. nike air max for sale
"What?" asked Beatrice, starting back, "Colonel Hall's----"
His lover's eye detected a wealth of cleverness in her correspondence. He sincerely believed that she was more gifted as a writer than himself. Her letters were full of descriptions of Fluffy in her part, thumb-nail sketches of the other members of the cast and accounts of the momentously personal adventures of a theatrical company on tour. She had a trick of humor that made her intimate in an adjective, and made him laugh. She also had a trick of allotting to him prejudices. "You'd call our leading man a very bad character, but I like him: I think one needs to have faults to be truly charitable. I'd ask you to join us, but—— You wouldn't get on with theatrical people; you rather—I know, so you needn't deny it—you rather despise them. I think they're the jolliest crowd. We dance every night when the show is ended and have late suppers, and—you can guess." CHAPTER XXI—THE EXPERIMENTAL HONEYMOON nike air max for sale "Haven't you a match?" "You were not near this place on that night?" They and the nation suffered accordingly. After the war, Lord Roberts embodied in a preface to the "Cavalry Training" Manual of 1904 the ripe experience, not only of the South African War, but of a long life spent in military service. He inculcated reliance on the rifle as the principal weapon for all purposes of the Cavalry soldier. Two years later, although Manchuria had confirmed his words in every particular, the injunction was forgotten, and our Cavalry were sitting at the feet of a German writer who had nothing to tell them about the rifle which they had not already learnt by costly war experience, and who was addressing, not them, but a Cavalry ignorant of the ABC of modern fire-tactics. But, as a matter of theory, not of experience, he clung to shock, expounding it in terms irreconcilable with fire. Our Manual of 1904 was superseded by the Manual of 1907, with the directions of Lord Roberts expunged and Bernhardi's self-contradictory counsels embodied. In the August number of the Revue des deux Mondes of 1908 many people were astonished to find set forth in full by General de Negrier, as a model to the "dreaming" French Cavalry, Lord Roberts's preface to our Manual of 1904. That Manual is cancelled. So that to find in its living, authoritative form the verdict of our greatest living soldier, derived from facts, not from theory, on a technical and tactical question of vital importance, the student has to search the files of a French review. nike air max for sale nike air max for sale "You are not complimentary, my friend," he said, turning away with a supercilious laugh. They stopped, Mr Verneede had hailed a cab, and Leavesley came out of his reverie to find a four-wheeler drawing up at the pavement. "You're eating my heart out, Desire. I'm good for nothing. You must say yes. If you don't love me, you at least like me. You like me immensely, don't you? The other will come later." His voice trembled with the need of her; it was more like crying. He tried to draw her to him; she clutched her wrap more tightly, and dodged across the threshold.
"Never clapped eyes on him in my life. A pretty woman belongs to the whole world, Teddy." "Good e'en to you, sir," he said, "you're looking piert and heartful. Have ye had good luck?" "And a good girl," said the old man, slightly warming. "She ain't ashamed of her old father, although she writes books and lives like a fine lady in London." "Now I'll wager," spoke up the Red Nosed Gentleman, "now I'll wager a bottle of burgundy—and that reminds me I must send for another, since this one by me is empty—that your experiences are quite as glorious as my own; and yet, sir,"—here the Red Nosed Gentleman looked hard at the Sour Gentleman as though defying him to the tiltyard—"should you favor us, I'll even follow you, and forage in the pages of my own heretofore and give you a story myself." CHAPTER VII—"PASHUN" IN THE KITCHEN She left him to place his own construction on the statement—to discover its origin in nobility or carelessness. Whichever it was, it made him the needle while she remained the magnet. When he wasn't with her, he was waiting for her; so the hours after midnight, when he had seen her home, were the only ones free from feverishness. His work suffered; he stole from the hours when he ought to have been in bed. He began to suspect that he was losing his confidence of touch. The suspicion was sharply confirmed when one of his commissioned articles came back with the cryptic intimation that it wasn't exactly what the editor had expected. That meant the loss of five hundred dollars; what was worse, it filled him with artistic panic.
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\section{Introduction: From soluble  to $\pi$-separable groups}

All groups considered are finite.

We pursue an extension of the theory of finite soluble groups to the universe of $\pi$-separable groups, $\pi$ a set of primes. Classical results of Hall theory state that soluble groups are characterized by the existence of Hall $\rho$-subgroups for all sets of primes $\rho$ (P. Hall~\cite{H28,H37}). When the set of primes $\pi$ is fixed, $\pi$-separable groups have Hall $\pi$-subgroups, and also every $\pi$-subgroup is contained in a conjugate of any Hall $\pi$-subgroup (S. A. \v{C}unihin~\cite{Cu}). The main results in this paper (Section~\ref{sCarter-like}, Theorems~\ref{teo1}~and \ref{teo2}) show that $\pi$-separable groups have a conjugacy class of subgroups which specialize to Carter subgroups, i.e. self-normalizing nilpotent  subgroups, or equivalently, nilpotent projectors, when specializing to soluble groups. In the theory of soluble groups, Carter subgroups are the cornerstone for the existence and conjugacy of injectors associated to Fitting classes. In a forthcoming paper \cite{ADFP}, our Carter-like subgroups are used to generalize these results to  $\pi$-separable groups.

With this aim we analyze first the reach of $\pi$-separability further from the universe of soluble groups. We refer to \cite{Go} for basic results on $\pi$-separable groups, and to \cite{DH} for background on classes of groups;  we shall adhere to their notations. If $\pi$ is a set  of primes, let us recall that a group $G$ is $\pi$-separable if every composition factor of $G$ is either a $\pi$-group or a $\pi'$-group, where $\pi'$ stands for the complement of $\pi$ in the set $\p$ of all prime numbers. We notice that a group $G$ is soluble if and only if it is $\rho$-separable for all sets of primes $\rho$. Regarding $\pi$-separability, $\pi$ a set of primes, it is clearly equivalent to $\pi'$-separability, so that there is no loss of generality to assume that $2\in \pi$. Then, by the Feit-Thompson theorem, a $\pi$-separable group is $\pi'$-soluble, i.e. the group is $\pi'$-separable with every $\pi'$-composition factor a $p$-group for some prime $p\in \pi'$. The following extension  for $\pi$-separable groups is  easily proved:

\begin{pro}\label{pro0} For a group $G$, if $2\in \pi\subseteq \p$ the following statements are pairwise equivalent:
\begin{enumerate} \item $G$ is $\pi$-separable;
\item $G$ is $\rho$-separable for every set of primes $\rho$ such that either $\pi \subseteq \rho$ or $\pi\cap \rho =\emptyset$;
\item $G$ is $\pi'$-separable ($\pi'$-soluble).
\end{enumerate}
\end{pro}

\begin{rem} The need of the hypothesis $2\in \pi$ for the validity of the equivalences in Proposition~\ref{pro0} is clear. Certainly, for a group $G$ and any set of primes $\pi$, it holds that  $2\rightarrow 1 \leftrightarrow 3$. But statement 2 implies that the $\pi'$-compositions factors of the group $G$ are soluble. Then, by the Feit-Thompson theorem, if $2\notin \pi$, statement 2 is equivalent to the solubility of the group $G$ and $1\nrightarrow 2$ in general.
\end{rem}

Consequently, by Proposition~\ref{pro0}, if $2\in \pi\subseteq \p$, every $\pi$-separable group possesses a Sylow $p$-complement of $G$, i.e. a Hall $p'$-subgroup of $G$, for each $p\in \pi'$, as well as a Hall $\pi'$-subgroup, and these  are pairwise permutable subgroups with coprime indices in the group. In analogy with \cite[I. Definitions (4.1), (4.5), (4.7)]{DH} it appears to be natural now to introduce the following concepts, which are proven to hold  in  $\pi$-separable groups if $2\in \pi$, by the previous comment and as explained below:

\begin{de} Let $G$ be a group and $\pi$ be a set of primes.
\begin{itemize}\item[$\mathbf{K}_\pi$:] A \emph{complement $\pi$-basis} of  $G$ is a set $\mathbf{K}_\pi$ containing exactly one Sylow $p$-complement of $G$, i.e. a Hall $p'$-subgroup of $G$, for each $p\in \pi'$, and exactly one Hall $\pi'$-subgroup.
\item[$\mathbf{\Sigma_\pi}$:] A \emph{Hall $\pi$-system} of  $G$ is a set $\mathbf{\Sigma_\pi}$ of Hall subgroups of $G$ satisfying the following two properties:
    \begin{enumerate}
    \item For each set of primes $\rho$ such that either $\pi \subseteq \rho$ or $\pi\cap \rho =\emptyset$, $\mathbf{\Sigma_\pi}$ contains exactly one Hall $\rho$-subgroup.
    \item If $H,K\in \mathbf{\Sigma_\pi}$, then $HK=KH$.
    \end{enumerate}
    \item[$\mathbf{B_\pi}$:] A \emph{Sylow $\pi$-basis} of  $G$ is a set $\mathbf{B_\pi}$ of subgroups of $G$ satisfying the following two properties:
    \begin{enumerate}
    \item $\mathbf{B_\pi}$  contains exactly one Hall $\pi$-subgroup, and exactly one Sylow $p$-subgroup for each $p\in \pi'$.
    \item If $H,K\in \mathbf{B_\pi}$, then $HK=KH$.
    \end{enumerate}
\end{itemize}
\end{de}

Obviously these systems may not exist in arbitrary  groups. By the previous comments, if the group is $\pi$-separable and $2\in \pi$, then complement $\pi$-bases do exist.

Note also that Sylow $\pi$-systems and  complement and Sylow  $\pi$-bases are hereditary with respect to normal subgroups and factor groups.

For any set $\rho$ of primes and a group $G$, we denote by $\Hall_{\rho}(G)$ the set of all Hall $\rho$-subgroups of $G$. If $p$ is a prime, then $\Syl_p(G)$ will denote the set of all Sylow $p$-subgroups of $G$. We keep mimicking the exposition in \cite[Section I.4]{DH}. The arguments there are easily adapted to prove the following corresponding results.

\begin{pro}\label{proKH}\textup{\cite[I. Proposition (4.4)]{DH}} Assume that the group $G$ has a complement $\pi$-basis, say $\mathbf{K_\pi}$ (particularly, if the group $G$ is  $\pi$-separable  and  $2\in \pi\subseteq \p$).  If $\rho$ is a set of primes such that $\pi\subseteq \rho$, let $G_\rho=\bigcap\{X\mid X\in \Hall_{p'}(G)\cap \mathbf{K_\pi},\ p\in \rho'\subseteq \pi'\}$. On the other hand, if $\rho$ is a set of primes such that $\pi\cap \rho=\emptyset$, let
$G_\rho=\bigcap\{X\mid (X\in \Hall_{p'}(G)\cap \mathbf{K_\pi},\ p\in \rho'\cap \pi')\ \vee\ (X\in \Hall_{\pi'}(G)\cap \mathbf{K_\pi})\}$. Then
\begin{enumerate}
\item $\mathbf{\Sigma_\pi}:= \{G_\rho\mid (\pi\subseteq \rho\subseteq \p)\ \vee (\rho \subseteq \p,\ \rho \cap \pi=\emptyset)\}$ is a Hall $\pi$-system of $G$, and
    \item $\mathbf{\Sigma_\pi}$ is the unique Hall $\pi$-system of $G$ containing $\mathbf{K_\pi}$.
\end{enumerate}

We shall say that $\mathbf{\Sigma_\pi}$ is the Hall $\pi$-system \emph{generated} by the complement $\pi$-basis $\mathbf{K_\pi}$.
\end{pro}

\begin{cor}\textup{\cite[I. Corollary (4.6)]{DH}} Let $G$ be a  $\pi$-separable group,  $2\in \pi\subseteq \p$. Then there is a bijective map between the set of all complement $\pi$-bases and the set of all Hall $\pi$-systems of $G$, such that to each complement $\pi$-basis corresponds the Hall $\pi$-system generated by it, and conversely, to each Hall $\pi$-system corresponds the complement $\pi$-basis contained in it.
\end{cor}

On the other hand, it is clear that every Hall $\pi$-system contains a unique Sylow $\pi$-basis. Also, each Sylow $\pi$-basis \emph{generates} a unique Hall $\pi$-system, by taking the product of the suitable elements in the basis to construct each element in the Hall $\pi$-system. We can easily state also the following result:

\begin{cor}\textup{\cite[I. Lemma (4.8)]{DH}} Let $G$ be a  $\pi$-separable group,  $2\in \pi\subseteq \p$. Then
there is a bijective map between the set of all Hall $\pi$-systems and the set of all Sylow $\pi$-bases of $G$, such that to each Hall $\pi$-system corresponds the Sylow $\pi$-basis contained in it, and conversely, to each Sylow $\pi$-basis corresponds the Hall $\pi$-system generated by it as described above.
\end{cor}

If $\mathbf{K_\pi}$ is a complement $\pi$-basis of a group $G$ and $g\in G$, it is clear that $\mathbf{K_\pi}^g:=\{X^g\mid X\in \mathbf{K_\pi}\}$ is again a complement $\pi$-basis of $G$, and this defines an action by conjugation of $G$ on the set of all complement $\pi$-bases of $G$. Analogously, any group $G$ acts by conjugation on the  set of all its Hall $\pi$-systems, and also on the set of all its Sylow $\pi$-bases.
\smallskip

By Proposition~\ref{pro0} again we have that if $G$ is a $\pi$-separable group, $2\in \pi\subseteq \p$, then $G$ acts transitively on the set of all Hall $p'$-subgroups for every $p\in \pi'$, as well as on the set of all Hall $\pi'$-subgroups, and the following result also holds.

\begin{teo}\textup{\cite[I. Theorems (4.9), (4.10), (4.11), Corollary (4.12)]{DH}} Let $G$ be a $\pi$-separable group, $2\in \pi$. Then:
 \begin{enumerate} \item The number of Hall $\pi$-systems of $G$ is $\prod_{S\in \mathbf{K_\pi}}|G:N_G(S)|$, where $\mathbf{K_\pi}$ is a complement $\pi$-basis of $G$.
 \item The group $G$ acts transitively by conjugation on the set of all complement $\pi$-bases, on the set of all Hall $\pi$-systems, as well as on the set of all Sylow $\pi$-bases.
     \end{enumerate}
\end{teo}

One might wish that the existence of Hall $\pi$-systems in finite groups would characterize $\pi$-separability, but this is not the case, even assuming the transitive action of the group by conjugation on the set of all Hall $\pi$-systems. The alternating group of degree $5$ together with the set $\pi=\{2,3\}$ is a counterexample. A characterization of $\pi$-separability by the existence of Hall subgroups had been in fact given by Z. Du~\cite{Du}, as shown in the next result. For notation, for a group $G$ and any set of primes $\rho$, the group $G$ is said to satisfy $E_{\rho}$ if $G$ has a Hall $\rho$-subgroup; if in addition each $\rho$-subgroup is contained in the conjugate of a Hall $\rho$-subgroup, it is said that $G$ satisfies $D_\rho$.

\begin{teo}\textup{\cite[Theorems 1, 3]{Du}} For a group $G$ and a set of primes $\pi$, the following statements are pairwise equivalent:
\begin{itemize}\item[(i)] $G$ is $\pi$-separable.
\item[(ii)] $G$ satisfies:
\begin{enumerate}
\item $E_\pi$ and $E_{\pi'}$;
\item $E_{\pi\cup \{q\}}$ and $E_{\pi'\cup \{p\}}$, for all $p\in \pi$, $q\in \pi'$.
\end{enumerate}
\item[(iii)] $G$ satisfies:
\begin{enumerate}
\item $E_\pi$ and $E_{\pi'}$;
\item $E_{\{p,q\}}$, for all $p\in \pi$, $q\in \pi'$.
\end{enumerate}
\end{itemize}
\end{teo}

We point out finally that the existence of Hall $\pi$-systems together with $\pi$-dominance do characterize $\pi$-separability, as we prove next.

\begin{rem}\label{dpidpi'} We notice that for any group $G$ and any $\pi\subseteq \p$, the existence of complement $\pi$-bases is equivalent to the existence of Hall $\pi$-systems, and also to the existence of Sylow $\pi$-bases, by Proposition~\ref{proKH} and the corresponding constructions. In this case the group $G$ satisfies $E_\pi$ and $E_{\pi'}$, and then, if $2\in \pi$, $G$ satisfies $D_{\pi'}$ (see \cite{AF}).
\end{rem}

For any positive integer $n$, we denote by $\pi(n)$ the set of primes dividing $n$; for  the order $|G|$ of a
group G, we set  $\pi(G)=\pi(|G|)$.

\begin{teo} Assume that the group $G$ has a complement $\pi$-basis and satisfies $D_\pi$, where $2\in \pi\subseteq \p$. Then $G$ is $\pi$-separable.
\end{teo}
\smallskip

{\it Proof.} We argue by induction on the order of $G$. Let $N$ be a normal subgroup
 of $G$. By \textup{(\cite[Theorem 7.7]{RV})}, $G$ satisfies $D_\pi$ if and only if $N$ and $G/N$ satisfies $D_\pi$. We may then assume that $G$ is a simple group. If $|\pi(G)\cap \pi'|\ge 2$,  then $G$ would satisfy $E_{p'}$ and $E_{q'}$, with $p,q$ odd different primes dividing the order of $G$, which would imply that $G$ would not be simple by \cite[Corollary 5.5]{AF}. Consequently we may assume that $|\pi(G)\cap \pi'|= 1$. On the other hand, by hypothesis and Remark~\ref{dpidpi'}, $G$ satisfies $D_\pi$ and $D_{\pi'}$, and we may assume that the group is neither a $\pi$-group nor a $\pi'$-group. In \cite[Lemma 3.1]{Gi}, such  a simple group is characterized to be  $G=PSL(2,q)$, where $q>3$, $q(q-1)\equiv 0(3)$, $q\equiv-1(4)$ and $\pi(q+1)\subseteq \pi$, $\pi(q(q-1)/2)\subseteq \pi'$. Hence $|\pi(G)\cap \pi'|\ge |\pi(q(q-1)/2)|\ge 2$, which is not possible and proves that $G$ is $\pi$-separable.\qed

\section{General framework: The theory of soluble groups}

Before focusing on the universe of $\pi$-separable groups, and proving our main results in Section~\ref{sCarter-like}, we present briefly here the general framework on the theory of soluble groups and classes of groups, where they are relevant to our concerns. It makes clear  the origin of main concepts in our extension to $\pi$-separable groups, $\pi$ a set of primes, particularly the one of $\mathfrak N^\pi$-Dnormal subgroups, associated to the class $\mathfrak N^\pi$ of groups which are the direct product of a $\pi$-group and a nilpotent $\pi'$-group.\smallskip

A Carter subgroup of a group is a self-normalizing nilpotent subgroup. The well-known result of R.~W.~Carter states that each finite soluble group possesses exactly one conjugacy class of Carter subgroups, namely the $\mathfrak  N$-projectors (or $\mathfrak  N$-covering subgroups), for the class $\mathfrak  N$ of finite nilpotent groups.
\smallskip

We recall that given a class of groups $\mathfrak  X$, a subgroup $U$ of a group $G$ is called an $\mathfrak  X$-projector of $G$ if $UK/K$ is an $\mathfrak  X$-maximal subgroup of $G/K$ (i.e. maximal as subgroup of $G/K$ in $\mathfrak  X$)  for all $K\unlhd G$. The (possibly empty) set of $\mathfrak  X$-projectors of $G$ will be denoted by $\proj_{\mathfrak  X}(G)$.

Also, an $\mathfrak  X$-covering subgroup of $G$ is a subgroup $E$ of $G$ with the property that $E\in \proj_{\mathfrak  X}(H)$ whenever $E\le H\le G$. The set of $\mathfrak  X$-covering subgroups of $G$ will be denoted by $\cov_{\mathfrak  X}(G)$.
\smallskip

Schunck classes  $\mathfrak  X$ appear as the classes of groups for which all finite groups have associated projectors.  If the group $G$ is soluble, then $\cov_{\mathfrak  X}(G)=\proj_{\mathfrak  X}(G)$  and it is a non-empty conjugacy class of subgroups of $G$. Saturated formations are classical and relevant special cases of Schunck classes.

 Let us recall also that a class of groups $\mathfrak  X$ is a formation if every epimorphic image of a group in $\mathfrak  X$ belongs to $\mathfrak  X$, and $G/(N_1\cap N_2)\in \mathfrak  X$ whenever $N_1,N_2\unlhd G$ with $G/N_1, G/N_2\in \mathfrak  X$. In this case,  the $\mathfrak  X$-residual of a group $G$, denoted $G^{\mathfrak  X}$, is the smallest normal subgroup of $G$ with quotient group in $\mathfrak  X$; for the class $\mathfrak  X=\mathfrak  E_\pi$ of all $\pi$-groups, $\pi$ a set of primes,  $G^{\mathfrak  E_\pi}=O^\pi(G)$ is the $\pi$-residual of $G$, also described as the subgroup generated by all $\pi'$-subgroups of $G$. A formation $\mathfrak  X$ is said to be saturated if $G\in \mathfrak X$ whenever $G/\Phi(G)\in \mathfrak X$, where $\Phi(G)$ denotes the Frattini subgroup of $G$.

 It is well-known that non-empty saturated formations are exactly local formations, i.e. classes of groups $LF(f)$ consisting of the groups $G$ such that,  for every prime $p\in \p$, it holds that $G/C_G(H/K)\in f(p)$ whenever $H/K$ is a chief factor of $G$ whose order is divisible by $p$, and where $f$ is a formation function, i.e. $f(p)$ is a (possibly empty) formation for each prime $p$. The characteristic of the saturated formation is $\Char (LF(f))=\{p\in \p\mid f(p)\neq \emptyset\}$, and $LF(f)$ is said to be locally defined by the formation function $f$. We shall refer also to the fact that a local formation $\mathfrak  F$ always has a smallest local definition, i.e. a formation function $\underline{f}$ such that $\mathfrak  F=LF(\underline{f})$, and $\underline{f}(p)\subseteq g(p)$ for every prime $p$ and any other formation function $g$ such that $\mathfrak  F=LF(\underline{f})=LF(g)$.

We refer to \cite[Chapters III, IV]{DH} for details related to the theories of Schunck classes and formations.
\smallskip

Let $\pi$ be a set of primes. Let $$\mathfrak N^\pi=\mathfrak  E_\pi \times \mathfrak  N_{\pi'}=(G=H\times K\mid H\in \mathfrak  E_\pi,\ K\in \mathfrak  N_{\pi'}),$$ $\mathfrak  E_\pi$  the class of all $\pi$-groups and $\mathfrak  N_{\pi'}$ the class of all nilpotent $\pi'$-groups.

In the particular cases when either $\pi=\emptyset$ or $\pi=\{p\}$, $p$ a prime, ($|\pi|\le 1$), then $\mathfrak N^\pi=\mathfrak  N$ is the class of all nilpotent groups.
\smallskip

Our main results Theorems~\ref{teo1}~and~\ref{teo2} extend the existence and properties of Carter subgroups in soluble groups to $\rho$-separable groups, $\rho$ a set of primes, with appropriate class $\mathfrak  N^\rho$ or $\mathfrak  N^{\rho'}$ playing the role of the class $\mathfrak  N$ of nilpotent groups.
\smallskip

We shall appeal also to the concept of  $\mathfrak N^\pi$-Dnormal subgroup, as $\mathfrak N^\pi$ is a saturated formation.
\smallskip

The concept of $\mathfrak  G$-Dnormal subgroups for a non-empty saturated formation $\mathfrak  G$, which was given by K. Doerk in the universe of finite soluble groups, and appears for the first time in \cite[Definition 3.1]{AP}, is also available for arbitrary finite groups, as defined next. For notation, if $G$ is a group, $\rho$ a set of primes, $G_\rho \in \Hall_\rho(G)$ and $H\le G$, we write $G_\rho\searrow H$ to mean that $G_\rho$ reduces into $H$, i.e. $G_\rho\cap H\in \Hall_\rho (H)$.

\begin{de}\textup{\cite[Definition 3.1]{AP}} Let $\mathfrak  G$ be a non-empty saturated formation and let $G$ be a group. A subgroup $H$ of $G$ is said to be $\mathfrak  G$-Dnormal in $G$ if $\pi(|G:H|)\subseteq \Char (\mathfrak  G)$, and for every $p\in \Char (\mathfrak  G)$ it holds that
$$[H_G^p,H^{\underline{g}(p)}]\le H,$$
where $\underline{g}$ denotes the smallest local definition of $\mathfrak  G$ as local formation, and $H_G^p=\langle G_p\in \Syl_p(G)\mid G_p\searrow H\rangle$.
\end{de}

The saturated formation $\mathfrak N^\pi=\mathfrak  E_\pi \times \mathfrak  N_{\pi'}=LF(f)=LF(\underline{f})$  is locally defined by the formation function $f$ given by  $f(p)=\mathfrak  E_p$ if $p\in \pi'$, and $f(p)=\mathfrak  E_\pi$ if $p\in \pi$. Then the smallest local definition is given by $\underline{f}(p)=(1)$ if $p\in \pi'$, and $\underline{f}(p)= \begin{cases}
(1)& \text{if }\pi=\{p\},\\
\mathfrak  E_\pi & \text{if }p\in \pi,\ |\pi|\ge 2.
\end{cases}$
\medskip

Hence, for $\mathfrak N^\pi=\mathfrak  N$, $\mathfrak  N$-Dnormal subgroups are exactly normal subgroups.
\smallskip

In the case $|\pi|\ge 2$, a subgroup $H$ of a group $G$ is $\mathfrak N^\pi$-Dnormal if it satisfies the following conditions:
\begin{description}
\item[$(1)$] whenever $p\in \pi'$ and $G_p\in \Syl_p(G)$, $G_p\searrow H$, then $G_p\le N_G(H)$;
\item[$(2)$] whenever $p\in \pi$ and $G_p\in \Syl_p(G)$, $G_p\searrow H$, then $G_p\le N_G(O^\pi(H))$.

\end{description}

Note that normal subgroups are $\mathfrak N^\pi$-Dnormal for any set of primes $\pi$.

\begin{rem} Regarding the previous statement, note that for any $X\le G$ it holds that $[X,O^\pi(H)]\le H$ if and only if $X\le N_G(O^\pi(H))$.
\smallskip

{\it Proof.} Assume that $[X,O^\pi(H)]\le H$. We consider $O^\pi(H)=\langle H_q\mid H_q\in \Syl_q(H),\ q\in \pi'\rangle$. Then
$[X,O^\pi(H)]\le \langle O^\pi(H)^x\mid x\in X\rangle=\langle H_q^x\mid x\in X,\ H_q\in \Syl_q(H),\ q\in \pi'\rangle=O^\pi(H).$ The converse is clear.\qed
\end{rem}

The next proposition provides a useful characterization of $\mathfrak N^\pi$-Dnormal subgroups.
\begin{pro}\label{pro1} Let $H$ be a subgroup  of a group $G$. Then:
\begin{enumerate}\item Assume that $|\pi|\le 1$. Then $\mathfrak N^\pi =\mathfrak  N$ and $H$ is $\mathfrak  N$-Dnormal in $G$ if and only $H$ is normal in $G$.
\item Assume that $|\pi|\ge 2$. Then the following statements are equivalent:
\begin{itemize} \item[(i)] $H$ is $\mathfrak N^\pi$-Dnormal in $G$;
\item[(ii)] $O^{\pi}(H)\unlhd G$ and $O^{\pi}(G)\le N_G(H)$.
\end{itemize}
\end{enumerate}
\end{pro}
   \smallskip

{\it Proof.} Part~1 is clear. For Part~2,  since $O^\pi(G)=\langle G_p\mid G_p\in \Syl_p(G),\ p\in \pi'\rangle$, it is clear that (ii) implies (i). Conversely, assume the (i) holds, i.e. $H$ is $\mathfrak N^\pi$-Dnormal in $G$. By Sylow's theorem, for each prime $p$, there exists $G_p\in \Syl_p(G)$ such that $G_p\searrow H$. The definition of $\mathfrak N^\pi$-Dnormality implies that $G_p\le N_G(O^{\pi}(H))$ if $p\in \pi$, and $G_p\le N_G(H)\le N_G(O^{\pi}(H))$ if $p\in \pi'$. Consequently,
$G=\langle G_p\mid p\in \p\rangle\le N_G(O^{\pi}(H))$, i.e. $O^{\pi}(H)\unlhd G$. In particular, for any $p\in \pi'$ and any $G_p\in \Syl_p(G)$, it holds that  $G_p\cap H=G_p\cap O^{\pi}(H)\in \Syl_p(O^{\pi}(H))=\Syl_p(H)$, which means that $G_p\searrow H$, and then $G_p\le N_G(H)$, because $H$ is $\mathfrak N^\pi$-Dnormal in $G$. Hence, $O^\pi(G)=\langle G_p\mid G_p\in \Syl_p(G),\ p\in \pi'\rangle\le N_G(H)$, and we are done.\qed
\medskip

For notation, whenever a group $X\in \mathfrak N^\pi$, we write $X=X_{\pi}\times X_{\pi'}$ where $X_{\pi}=O_{\pi}(X)\in \mathfrak  E_{\pi}$ and $X_{\pi'}=O_{\pi'}(X)\in \mathfrak  N_{\pi'}$.
\smallskip

\begin{cor}\label{cor1} Assume that $|\pi|\ge 2$ and let $H$ be a subgroup of a group $G$ such that $H=H_\pi\times H_{\pi'}\in \mathfrak N^\pi$.
Then $H$ is $\mathfrak N^\pi$-Dnormal in $G$ if and only if $H_{\pi'}\unlhd G$ and $O^{\pi}(G)\le N_G(H)$.
\end{cor}
\smallskip

{\it Proof.} This is a consequence of Proposition~\ref{pro1}(2) since in this case $H_{\pi'}=O^{\pi}(H)$.\qed

\section{Carter-like subgroups in $\pi$-separable groups}\label{sCarter-like}

Let $\pi$ be a set of primes.  As above, set $\mathfrak N^\pi=\mathfrak  E_\pi \times \mathfrak  N_{\pi'}$, where $\mathfrak  E_\pi$ is the class of $\pi$-groups and $\mathfrak  N_{\pi'}$ is  the class of nilpotent $\pi'$-groups.
 \smallskip

 We prove in this section that if $G$ is a $\pi'$-soluble group, then $\mathfrak N^\pi$-projectors coincide with $\mathfrak N^\pi$-covering subgroups, and they form a conjugacy class of self-$\mathfrak N^\pi$-Dnormalizing subgroups of $G$ (see Definition~\ref{defselfnormal}, and  Theorems~\ref{teo1}, \ref{teo2}). If $\mathfrak N^\pi=\mathfrak  N$ is the class of nilpotent groups and $G$ is a soluble group, these are the Carter subgroups.
 \smallskip

We notice that Burnside's $p^aq^b$-theorem together with the Feit-Thompson theorem imply that $\pi$-separable groups are $\pi'$-soluble whenever  $|\pi'|\le 2$, or $2\in \pi$ if $|\pi'|\ge 3$.
\smallskip

Notice also that, by the Feit-Thompson theorem, for any set of primes $\rho$, a $\rho$-separable group is either $\rho$-soluble or $\rho'$-soluble.
\smallskip

For our main results we quote a series of results from \cite[III. Section 3]{DH} and adhere to the notation there, though specialized  to our saturated formation $\mathfrak N^\pi$ and our purposes.  We notice that  $b(\mathfrak N^\pi)\subseteq \mathcal P_1\cup\mathcal P_2$, where $b(\mathfrak N^\pi)$ is the class of groups $G\notin \mathfrak N^\pi$ but whose proper epimorphic images belong to $\mathfrak N^\pi$, and for each $i=1,2$, $\mathcal P_i$ is the class of primitive groups with a unique minimal normal subgroup, which is abelian for $i=1$, and non-abelian for $i=2$.

\begin{lem}\label{lem2} \begin{enumerate} \item \emph{\cite[III. Proposition (3.7)]{DH}} For a group $G$, whenever $N\unlhd G$, $N\le V\le G$, $U\in \proj_{\mathfrak N^\pi}(V)$, and $V/N\in \proj_{\mathfrak N^\pi}(G/N)$, then $U\in \proj_{\mathfrak N^\pi}(G)$.
\item \emph{\cite[III. Lemma (3.9)]{DH}} Assume that $G\in b(\mathfrak N^\pi)$. Then:
\begin{enumerate}\item If $G\in \mathcal P_1$, then $\cov_{\mathfrak N^\pi}(G)$ and $\proj_{\mathfrak N^\pi}(G)$ both coincide with the non-empty set comprising those subgroups of $G$ which are complements in $G$ to the minimal normal subgroup of $G$.
\item If $G\in \mathcal P_2$, then $\proj_{\mathfrak N^\pi}(G)$ is non-empty and consists of all $\mathfrak N^\pi$-maximal subgroups of $G$ which supplement the minimal normal subgroup of $G$ in $G$.
\end{enumerate}

\item \emph{\cite[III. Theorem (3.10)]{DH}} For any group $G$, $\cov_{\mathfrak N^\pi}(G)\subseteq \proj_{\mathfrak N^\pi}(G)\neq \emptyset$.

\item \emph{\cite[III. Theorem (3.14)]{DH}} Let $N$ be a nilpotent normal subgroup of a group $G$, and let $H$ be an $\mathfrak N^\pi$-maximal subgroup of $G$ such that $G=HN$. Then $H\in \proj_{\mathfrak N^\pi}(G)$.

\item \emph{(\cite[III. Theorem (3.19)]{DH},\cite{F})} Let $\mathfrak B$ be the class of all $\pi'$-soluble groups. The statement ``\,$\proj_{\mathfrak N^\pi}(G)$ is a conjugacy class of $G$'' is true for all groups $G\in \mathfrak B$ if and only if it is true for all $G\in b(\mathfrak N^\pi)\cap \mathfrak B$.
    \item \emph{(\cite[III. Remark (3.20)(b)]{DH},\cite{F})} Let $\mathfrak B$ be the class of all $\pi'$-soluble groups. The statement ``\,$\proj_{\mathfrak N^\pi}(G)=\cov_{\mathfrak N^\pi}(G)$'' is true for all groups $G\in \mathfrak B$ if and only if it is true for all $G\in b(\mathfrak N^\pi)\cap \mathfrak B$.
\end{enumerate}
\end{lem}

We still quote the following result for our purposes.

\begin{lem}\label{lem2+}\emph{\cite[Theorems 4.1.18, 4.2.17]{BE}} Let $\mathfrak  H$ be a saturated formation and let $G$ be a group  whose $\mathfrak  H$-residual $G^{\mathfrak  H}$ is abelian. Then $G^{\mathfrak  H}$ is complemented in $G$, any two complements are conjugate in $G$, and the complements are the $\mathfrak  H$-projectors of $G$.
\end{lem}

\begin{lem}\label{lem3} Let $M=M_{\pi}\times M_{\pi'}$ be an $\mathfrak N^\pi$-maximal subgroup of a $\pi$-separable group $G$. Then:
\begin{enumerate}\item $M=M_{\pi'}C_G(M_{\pi'})_{\pi}$ for some $C_G(M_{\pi'})_{\pi}\in Hall_{\pi} (C_G(M_{\pi'}))$.
\item If $H=H_{\pi}\times H_{\pi'}$ is another $\mathfrak N^\pi$-maximal subgroup of $G$ and $M_{\pi'}^x= H_{\pi'}$ for some $x\in G$, then $M^g=H$ for some $g\in G$.
\end{enumerate}
\end{lem}

\smallskip

{\it Proof.} 1. We have that $M_{\pi}\le C_G(M_{\pi'})$ and so $M_{\pi}\le C_G(M_{\pi'})_{\pi}$ for some $C_G(M_{\pi'})_{\pi}\in Hall_{\pi} (C_G(M_{\pi'}))$. But $M_{\pi'}C_G(M_{\pi'})_{\pi}\in \mathfrak N^\pi$, which implies that  $M=M_{\pi'}C_G(M_{\pi'})_{\pi}\in \mathfrak N^\pi$ by the maximality of $M$.
\smallskip

2. The hypothesis implies that $C_G(M_{\pi'})^x= C_G(M_{\pi'}^x) =
C_G(H_{\pi'}) $. Then $M_{\pi}^x, H_{\pi}\in \Hall_{\pi}(C_G(H_{\pi'}))$ and $M_{\pi}^{xy}=H_{\pi}$ for some $y\in C_G(H_{\pi'})$. Consequently,
$M^{xy}= M_{\pi}^{xy}M_{\pi'}^{xy}=H_{\pi}H_{\pi'}=H$, and we are done.\qed

\begin{teo}\label{teo1} If $G$ is a $\pi'$-soluble group, then $\emptyset \neq \proj_{\mathfrak N^\pi}(G)= \cov_{\mathfrak N^\pi}(G)$ and it is a conjugacy class of $G$.
\end{teo}
\smallskip

{\it Proof.} By Lemma~\ref{lem2}, parts~(3),~(5),~(6), we may assume that $G\in b(\mathfrak N^\pi)$. Since $\mathfrak N^\pi$ is a saturated formation, $G\in \mathcal P_1\cup \mathcal P_2$; let $N$ be the minimal normal subgroup of $G$.

If $G\in \mathcal P_1$, the result follows by Lemmas~\ref{lem2}(2)(a)~and~\ref{lem2+}.

Assume now that $G\in \mathcal P_2$.  We know by Lemma~\ref{lem2}(2)(b) that $\proj_{\mathfrak N^\pi}(G)$ is non-empty and consists of all $\mathfrak N^\pi$-maximal subgroups of $G$ which supplement $N$ in $G$. We prove first that these subgroups are conjugate in $G$.

Let $M=M_{\pi}\times M_{\pi'}$, $H=H_{\pi}\times H_{\pi'}$ be $\mathfrak N^\pi$-maximal subgroups of $G$ such that $G=NM=NH$. Since $G$ is $\pi'$-soluble,  and $N$ is non-abelian, $N$ is a $\pi$-group and, consequently, $M_{\pi'}, H_{\pi'}\in \Hall_{\pi'}(G)$. Hence, there exists $x\in G$ such that $M_{\pi'}^x=H_{\pi'}$ and $M$ and $H$ are conjugate by Lemma~\ref{lem3}(2).
\smallskip

We claim now that $\proj_{\mathfrak N^\pi}(G)\subseteq \cov_{\mathfrak N^\pi}(G)$, which will conclude the proof.

Let $M\in \proj_{\mathfrak N^\pi}(G)$, i.e. $M=M_{\pi}\times M_{\pi'}$ is an $\mathfrak N^\pi$-maximal subgroup of $G$ such that $G=NM$. Let $M\le L\le G$. We aim to prove that $M\in \proj_{\mathfrak N^\pi}(L)$. Let $T\in \proj_{\mathfrak N^\pi}(L)$. We notice that $L=M(L\cap N)$. Then $L/(L\cap N)\cong M/(M\cap N)\in \mathfrak N^\pi$. Since $T(L\cap N)/(L\cap N)$ is  $\mathfrak N^\pi$-maximal in $L/(L\cap N)$, it follows that $L=T(L\cap N)$. Hence $T_{\pi'}, M_{\pi'}\in \Hall_{\pi'}(L)$ and, moreover, $T$ and $M$ are $\mathfrak N^\pi$-maximal subgroups of $L$. By Lemma~\ref{lem3}(2), $T$ and $M$ are conjugate in $L$ and $M\in \proj_{\mathfrak N^\pi}(L)$.\qed

\begin{rem} In Theorem~\ref{teo1} the hypothesis of $\pi'$-solubility cannot be weakened to $\pi$-separability.  Otherwise, for the particular case when $\pi=\emptyset$, the result would hold for every finite group and the formation $\mathfrak N^\pi=\mathfrak  N$ of nilpotent groups, which is not true. Particularly, also if $\pi\neq \emptyset$, one can consider for instance $\pi=\p - \{2,3,5\}$, $\pi'=\{2,3,5\}$ and $G=\Alt(5)$ the alternating group of degree $5$. The group $G$ is obviously $\pi$-separable, the $\mathfrak  N^{\pi}$-projectors are the $\mathfrak  N$-projectors, which do not form a conjugacy class of subgroups, as they are all the Sylow subgroups of $G$; and $G$ has no $\mathfrak  N$-covering subgroups.
\end{rem}

\begin{de}\label{defselfnormal}
A subgroup $H$ of a group $G$ is said to be \emph{self-$\mathfrak N^\pi$-Dnormalizing} in $G$ if whenever $H\le K\le G$ and $H$ is $\mathfrak N^\pi$-Dnormal in $K$, then $H=K$.
\end{de}

We prove next that $\mathfrak N^\pi$-projectors are self-$\mathfrak N^\pi$-Dnormalizing subgroups.

\begin{pro}\label{prop2} Let $H$ be an $\mathfrak N^\pi$-projector of a  $\pi'$-soluble group $G$. Then $H$ is self-$\mathfrak N^\pi$-Dnormalizing in $G$.
\end{pro}
\smallskip

{\it Proof.} Assume that $H\le K\le G$ and $H$ is $\mathfrak N^\pi$-Dnormal in $K$. We aim to prove that $H=K$. By Theorem~\ref{teo1} and Corollary~\ref{cor1}, we have that $H\in \cov_{\mathfrak N^\pi}(K)$ and $H_{\pi'}\unlhd K$. Then $H/H_{\pi'}\in \proj_{\mathfrak N^\pi}(K/H_{\pi'})$   and $H/H_{\pi'}\le K_{\pi}H_{\pi'}/H_{\pi'}\in \mathfrak N^\pi$, for any $K_{\pi}\in \Hall_{\pi}(K)$, which implies that $H=K_{\pi}H_{\pi'}$. Whence, if $K_{\pi'}\in \Hall_{\pi'}(K)$, then $K=HK_{\pi'}$ and $H\unlhd K$ by Proposition~\ref{pro1}. If $H< K$, then $H< HK_p$ for some $p\in \pi'$ and $1\ne K_p\in \Syl_p(K)$. But $HK_p/H\in \mathfrak N^\pi$ which contradicts the fact that $H\in \cov_{\mathfrak N^\pi}(K)$.\qed

\begin{lem}\label{lem4} Assume that $|\pi|\ge 2$, $H=H_{\pi}\times H_{\pi'},K=K_{\pi}\times K_{\pi'}\in \mathfrak N^\pi$ and $H\le K$. Then $H$ is $\mathfrak N^\pi$-Dnormal in $K$ if and only if $H_{\pi'}\unlhd K_{\pi'}$.
\end{lem}
\smallskip

{\it Proof.} We notice that $[K_{\pi},H_{\pi'}] =1$ and $[K_{\pi'},H_{\pi}] =1$. Consequently, Corollary~\ref{cor1} implies that $H$ is $\mathfrak N^\pi$-Dnormal in $K$ if and only if $H_{\pi'}\unlhd K_{\pi'}$.\qed

\begin{pro} Assume that $H=H_{\pi}\times H_{\pi'}< L=L_{\pi}\times L_{\pi'}\in \mathfrak N^\pi$. Then there exists $K\le L$ such that $H<K$ and $H$ is $\mathfrak N^\pi$-Dnormal in $K$.
\end{pro}
\smallskip

{\it Proof.} If $|\pi|\le 1$, then $\mathfrak N^\pi = \mathfrak  N$, and the result is clear. In the case $|\pi|\ge 2$, by Lemma~\ref{lem4}, if $H_{\pi'}=L_{\pi'}$, then $H$ is $\mathfrak N^\pi$-Dnormal in $L$, and we are done. Otherwise, there exists $T\le L_{\pi'}$ such that $H_{\pi'}\vartriangleleft T$, i.e. $H_{\pi'}$ is a proper normal subgroup of $T$, because $L_{\pi'}$ is nilpotent. We can consider now the subgroup $K=L_{\pi}T\le L$ which satisfies that $H<K$ and $H$ is $\mathfrak N^\pi$-Dnormal in $K$, which concludes the proof.\qed
\smallskip

As a consequence we can state the following:

\begin{cor}\label{cor2} If $H\in \mathfrak N^\pi$ is a self-$\mathfrak N^\pi$-Dnormalizing subgroup of a group $G$, then $H$ is $\mathfrak N^\pi$-maximal in $G$.
\end{cor}
\begin{rem} It is not true in general that $\mathfrak N^\pi$-projectors  of $\pi'$-soluble groups are exactly self-$\mathfrak N^\pi$-Dnormalizing subgroups in $\mathfrak N^\pi$. Otherwise,  $\mathfrak N^\pi\cap \mathfrak  S$ would be either $\mathfrak  N$ or $\mathfrak  S$, the class of all soluble groups, by \cite[Proposition 4.1]{AP04}. But we see next that a corresponding result to \cite[Theorem 4.2]{AP04} is still possible. That reference provides a corresponding result to our next Theorem~\ref{teo2}, for finite soluble groups, subgroup-closed saturated formations and associated projectors.
\end{rem}

As a consequence of Lemma~\ref{lem2+} we can state the following.

\begin{lem}\label{lem5} Let $\mathfrak  H$ be a saturated formation, $X$ be a group and $H$ be an $\mathfrak  H$-projector of $X$. Then $H\cap X^{\mathfrak  H}\le (X^{\mathfrak  H})'$.
\end{lem}

\begin{teo}\label{teo2}  For a subgroup $H$ of a $\pi'$-soluble group $G$ the following statements are pairwise equivalent:
\begin{enumerate}\item $H$ is an $\mathfrak N^\pi$-projector of $G$.
\item $H$ is an $\mathfrak N^\pi$-covering subgroup of $G$.
\item $H\in \mathfrak N^\pi$ is a self-$\mathfrak N^\pi$-Dnormalizing subgroup of $G$ and $H$ satisfies the following property:
\begin{gather}\text{If}\  H\le X\le G,\ \text{then}\ H\cap X^{\mathfrak N^\pi}\le (X^{\mathfrak N^\pi})'.\tag{*}
\end{gather}

\end{enumerate}
\end{teo}
\smallskip

{\it Proof.} The equivalence $1 \leftrightarrow 2$ has been proven in Theorem~\ref{teo1}. On the other hand, Proposition~\ref{prop2} and Lemma~\ref{lem5} prove $2\rightarrow 3$. We prove next that $3\rightarrow 1$.
\smallskip

Let  $H\in \mathfrak N^\pi$ be a self-$\mathfrak N^\pi$-Dnormalizing subgroup of $G$ satisfying property $(^*)$. We aim to prove that $H\in \proj_{\mathfrak N^\pi}(G)$. We notice that $H$ is $\mathfrak N^\pi$-maximal in $G$ by Corollary~\ref{cor2}.
\smallskip

If $G\in \mathfrak N^\pi$, then $H=G$ and the result follows. So that we may assume that $G\notin \mathfrak N^\pi$. We argue by induction on the order of $G$. Let $N$ be a minimal normal subgroup of $G$ such that $N\le G^{\mathfrak N^\pi}$.
\smallskip

We distinguish the following cases:
\begin{description}
\item[Case 1.]  $G=HN$.
\item[Case 2.] $HN<G$.
\end{description}
\smallskip

\noindent
{\bf Case 1.}  If $N$ is abelian, the result follows by Lemma~\ref{lem2}(4). Assume that $N$ is not abelian. Let $K\unlhd G$ such that $G/K\in b(\mathfrak N^\pi)$. Then $N$ is not contained in $K$ because $G/N\cong H/(H\cap N)\in \mathfrak N^\pi$. In particular, $N\cap K=1$ and $N\cong NK/K$ is a minimal normal subgroup of $G/K=(NK/K)(HK/K)\in \mathcal P_2$, with $HK/K<G/K$ because $HK/K\in \mathfrak N^\pi$. By Lemma~\ref{lem2}(2)(b), $HK/K\le P/K$ for some $P/K\in \proj_{\mathfrak N^\pi}(G/K)$.
We have now that $H\le P<G$. The inductive hypothesis implies that $H\in \proj_{\mathfrak N^\pi}(P)$, and from Lemma~\ref{lem2}(1), $H\in \proj_{\mathfrak N^\pi}(G)$, as claimed.
\smallskip

\noindent
{\bf Case 2.}  In this case $HN<G$ and the inductive hypothesis implies that $H\in \proj_{\mathfrak N^\pi}(HN)$. We prove first that $HN/N$ satisfies property $(^*)$ in $G/N$. Assume that $HN/N\le X/N\le G/N$. If $X<G$, then $H\in \proj_{\mathfrak N^\pi}(X)$ by inductive hypothesis and then $HN/N\cap (X/N)^{\mathfrak N^\pi}\le ((X/N)^{\mathfrak N^\pi})'$.
Otherwise, $X=G$ and so $(HN/N)\cap (X/N)^{\mathfrak N^\pi}= (HN/N)\cap (G/N)^{\mathfrak N^\pi}= (HN/N)\cap G^{\mathfrak N^\pi}/N= (H\cap G^{\mathfrak N^\pi})N/N\le (G^{\mathfrak N^\pi})'N/N= ((G/N)^{\mathfrak N^\pi})'=((X/N)^{\mathfrak N^\pi})'.$

We claim that $HN/N\in \mathfrak N^\pi$ is self-$\mathfrak N^\pi$-Dnormalizing in $G/N$. Then the result follows by inductive hypothesis together with Lemma~\ref{lem2}(1).
\smallskip

Assume that $HN/N$ is $\mathfrak N^\pi$-Dnormal in $L/N\le G/N$.

If $L<G$, the inductive hypothesis implies that $H$ is an $\mathfrak N^\pi$-projector of $L$ and so $HN=L$ by Proposition~\ref{prop2}. So that we may assume that $L=G$ and $HN/N$ is $\mathfrak N^\pi$-Dnormal in $ G/N$.
\smallskip

We split the rest of the proof into the following steps:
\smallskip

\noindent
{\it Step 1.} $N_G(HN)=HN$.
\smallskip

If $g\in N_G(HN)$, then $HN=H^gN$. Since $H\in \proj_{\mathfrak N^\pi}(HN)$, and this is a conjugacy class of subgroups of $HN$ by Theorem~\ref{teo1}, $H^g=H^x$ for some $x\in HN$. Consequently, $gx^{-1}\in N_G(H)=H$, and $g\in HN$.
\smallskip

\noindent
{\it Step 2.} $O^{\pi}(G)\le HN$.
\smallskip

Since $HN/N$ is $\mathfrak N^\pi$-Dnormal in $G/N$, we have by Proposition~\ref{pro1} and Step~1 that $O^{\pi}(G)N/N\le N_{G/N}(HN/N)=HN/N$.
\smallskip

\noindent
{\it Step 3.} $O^{\pi}(G)H=HN$ and it is a maximal subgroup of $G$.
\smallskip

Assume that $O^{\pi}(G)\le HN\le T< G$. The inductive hypothesis implies that $H\in \proj_{\mathfrak N^\pi}(T)$. Moreover, $T/O^{\pi}(G)\in \mathfrak  E_{\pi}\subseteq \mathfrak N^\pi$. Hence $T=O^{\pi}(G)H=HN$.
\smallskip

\noindent
{\it Step 4.} $G/N= (H_{\pi'}N/N)\times O_{\pi}(G/N)\in \mathfrak N^\pi$. In particular, $G^{\mathfrak N^\pi}=N$.
\smallskip

We have that $H_{\pi'}N/N\unlhd G/N$ and $H_{\pi'}N/N\in \Hall_{\pi'}(G/N)$ by Corollary~\ref{cor1} and Step~2. Since $[H_{\pi},H_{\pi'}]=1$ and $G/N$ is $\pi$-separable, and equivalently $\pi'$-separable, it follows  by \cite[6. Theorem 3.2]{Go} that \begin{gather*}(H_\pi N/N)O_\pi (G/N)/O_\pi (G/N)\le C_{(G/N)/O_\pi(G/N)}(O_{\pi'}((G/N)/O_\pi(G/N)))\\ \le O_{\pi'}((G/N)/O_\pi(G/N)),\end{gather*} which implies  $H_{\pi}N/N\le O_{\pi}(G/N)$.

If $H_{\pi}N/N= O_{\pi}(G/N)$, then $HN/N\unlhd G/N$ and $G=HN$ by Step~1, a contradiction.

Consequently, we may assume that $H_{\pi}N/N< O_{\pi}(G/N)$. Then $HN/N< (H_{\pi'}N/N) O_{\pi}(G/N)\le G/N$. By Step~3, $(H_{\pi'}N/N) O_{\pi}(G/N)= G/N \in \mathfrak N^\pi$.
\smallskip

\noindent
{\it Step 5.}  $G=NP$ where $P=P_\pi\times P_{\pi'}\in \proj_{\mathfrak N^\pi}(G)$.
\smallskip

It follows by Step~4.
\smallskip

\noindent
{\it Step 6.}  $N\in \mathfrak  E_{\pi'}$.
\smallskip

If $N\in \mathfrak  E_{\pi}$, then $H_{\pi'}, P_{\pi'}\in \Hall_{\pi'}(G)$. Since $H$ and $P$ are both $\mathfrak N^\pi$-maximal subgroups, it follows by Lemma~\ref{lem3} that $H=P^x\in \proj_{\mathfrak N^\pi}(G)$ and then $HN=G$, a contradiction. Since $G$ is $\pi$-separable, $N\in \mathfrak  E_{\pi'}$.
\smallskip

\noindent
{\it Step 7.}  Final contradiction.
\smallskip

Steps~4,~6, imply that $N=G^{\mathfrak N^\pi}$ is abelian, since $G$ is $\pi'$-soluble. By hypothesis, $H\cap N=1$. Moreover, $P_{\pi'}N=H_{\pi'}N= O_{\pi'}(G)$ and $P_\pi\in \Hall_{\pi}(G)$. There is no loss of generality to assume that $H_\pi\le P_{\pi}$ and then $[H_\pi, P_{\pi'}]=1$.

Since $H$ is an $\mathfrak N^\pi$-maximal subgroup of $HN=H_{\pi}H_{\pi'}N$, we deduce that $H_{\pi'}$ is an $\mathfrak  N$-maximal subgroup of $C_{H_{\pi'}N}(H_{\pi})=H_{\pi'}C_N(H_{\pi})$. But $C_N(H_{\pi})\unlhd C_{H_{\pi'}N}(H_{\pi})$ and $C_N(H_{\pi})$ is nilpotent. By Lemma~\ref{lem2}(4), $H_{\pi'}\in \proj_{\mathfrak  N}(C_{H_{\pi'}N}(H_{\pi}))$. On the other hand, $$C_{H_{\pi'}N}(H_{\pi})=C_{P_{\pi'}N}(H_{\pi})=P_{\pi'}C_{N}(H_{\pi}).$$ Since $P_{\pi'}$ is nilpotent, it follows again by Lemma~\ref{lem2}(4), that $P_{\pi'}\le H_{\pi'}^x\in \proj_{\mathfrak  N}(C_{H_{\pi'}N}(H_{\pi}))$, for some $x\in C_N(H_\pi)$. If
$P_{\pi'}< H_{\pi'}^x$, since $P_{\pi'}N=H_{\pi'}^xN$, we have that $H_{\pi'}^x\cap N\neq 1$ and so also $H\cap N\neq 1$, a contradiction. Therefore,
$P_{\pi'}= H_{\pi'}^x$, and Lemma~\ref{lem3} implies that $H^g=P$ for some $g\in G$ as $H$ and $P$ are $\mathfrak N^\pi$-maximal subgroups of $G$. It follows that $H\in \proj_{\mathfrak N^\pi}(G)$ and $HN=G$, the final contradiction.\qed

\bigskip

\noindent
{\bf Acknowledgments.} Research supported by Proyectos PROMETEO/2017/ 057 from the Generalitat Valenciana (Valencian Community, Spain), and PGC2018-096872-B-I00 from the Ministerio de Ciencia, Innovaci\'on y Universidades, Spain, and FEDER, European Union. The fourth author acknowledges with thanks the financial support of the Universitat de Val\`encia as research visitor (Programa Propi d'Ajudes a la Investigaci\'o de la Universitat de Val\`encia, Subprograma Atracci\'o de Talent de VLC-Campus, Estades d'investigadors convidats (2019)).

redpajama_set_name: RedPajamaArXiv


text:
"""
Written by Jesse Evers

Finds the distance between two cities.
Allows the user to specify whether they want miles or kilometers.
"""

import math

# Gets the latitude and longitude of the cities
def cities():

    city1 = input("Where would you like to start? ")
    city2 = input("Where would you like to end? ")

    lat1 = float(input("Enter the latitude of " + city1 + " in decimal format (negative for south): "))
    long1 = float(input("Enter the longitude of " + city1 + " in decimal format (negative for west: "))
    lat2 = float(input("Enter the latitude of " + city2 + " in decimal format (negative for south): "))
    long2 = float(input("Enter the longitude of " + city2 + " in decimal format (negative for west): "))
    calcDistance(lat1, long1, lat2, long2)

# Finds the distance between (lat1, long1) and (lat2, long2)
def calcDistance(lat1, long1, lat2, long2):

    radius = 3963  # Radius of the earth

    # Latitude and longitude measurements in radians
    lat1Rad = math.radians(lat1)
    lat2Rad = math.radians(lat2)
    long1Rad = math.radians(long1)
    long2Rad = math.radians(long2)

    # This is the haversin formula for calculating distances with latitude and longitude.
    # I don't really understand it, but it uses the haversine function, which is defined later.
    # See http://en.wikipedia.org/wiki/Haversine_formula for more details.
    distance = 2 * radius * math.asin(math.sqrt(haversin(lat2Rad - lat1Rad) + math.cos(lat1Rad) * math.cos(lat2Rad) * haversin(long2Rad - long1Rad)))

    # Decides on miles or kilometers
    units = int(input("Enter 1 for miles, 2 for kilometers: "))

    if units == 2:
        distance *= 1.60934
    elif units != 1:
        print("Not a valid choice, displaying in miles...")

    print(str.format("{0:.2f}", distance))

# Defines the haversine function
def haversin(theta):

    return pow(math.sin(theta / 2), 2)

redpajama_set_name: RedPajamaGithub


Haemochromatosis is the most common inherited disorder that causes the body to retain excessive amounts of iron.1 It is ten times more common in males and results in iron accumulation in various organs, in particular the liver and pancreas. The prevalence in various northern European populations is estimated to be as high as 1 in 200.
The relationship between haemochromatosis and diabetes mellitus has been well established and documented in medical literature, hence the term 'bronze diabetes'. Diabetes affects 30% to 60% of patients with hereditary haemochromatosis.2 Although the underlying pathophysiology of diabetes in patients with haemochromatosis has not been fully elucidated, it is thought to be multifactorial.
A 60 year old male patient with a background of epilepsy presented to his primary care physician in 2003 with an episode of candida balanitis. He had a two month history of osmotic symptoms and two stone weight loss. His random blood glucose was 21.2mmol/L. Although he was not ketoacidotic and his weight was 70kg, given the brief history, he was diagnosed with type 1 diabetes. He started twice daily mix insulin regime and remained relatively well controlled with HbA1c ranging between 6.5% and 7.7% and free of diabetes related complications. He had no family history of diabetes and his drug history included phenytoin 200mg bd.
At the time of presentation he was noted to have elevated transaminases: AST 49 IU/L (reference range 0–38 IU/L) and ALT 69 IU/L (reference range 0–38 IU/L). He drank about 8 units of alcohol per week and had no risk factors for hepatitis. Abdominal ultrasound showed a slightly enlarged and echogenic liver which was thought to be secondary to fatty infiltration. Monitoring of liver function continued for 6 years and unfortunately no definitive diagnosis was made.
In 2009, a comprehensive liver screen was requested as shown in table 1. This confirmed a hugely elevated ferritin level, thus raising the possibility of haemochromatosis.
Subsequent genetic testing for the C282Y gene confirmed type 1 hereditary haemochromatosis. He was referred to the gastroenterology team and commenced treatment with weekly venesections. Repeated ultrasound scans were negative for hepatic cirrhosis.
In February 2013 the patient had a hospital admission for labyrinthitis and was found to have had recurrent hypoglycaemic episodes. He seemed to have lost his hypo-awareness and was observed to have very tight control with an HbA1c of 5.6% (38mmol/mol). He was referred to the endocrinology team and as a new patient we reviewed his notes and found that his deranged liver function tests actually dated back to 1996, which in fact pre-dated his diabetes. It was only then thought that his diabetes may have been secondary to his undiagnosed haemochromatosis rather than type 1 diabetes (See Table 2 for trend of liver function tests, ferritin and HbA1c from 2003 onwards).
He was still on insulin therapy and had no documented ketoacidotic episodes. There were no reports of him having missed insulin doses with any undue effects. To further clarify his diabetes diagnosis, GADA were requested. This was negative at <5 IU/mL (reference range 0-10 IU/mL). Anti-islet cell antibodies were not done.
Baseline pituitary function tests and a short synacthen test were requested in view of his recurrent hypoglycaemic episodes (see Table 3). On further questioning he was also experiencing low energy, loss of libido and erectile dysfunction. The patient was found to have hypogonadotrophic-hypogonadism secondary to his haemochromatosis. He was started on six-weekly testosterone injections with good effect. He also had an MRI scan of his pituitary which was normal.
9. McDermott JH, Walsh CH. Hypogonadism in hereditary haemochromatosis. JCEM 2005;90:2451-5.
10. Feldman HA, Longcope C, Derby C, et al. Age trends in the level of serum testosterone and other hormones in middle-aged men: longitudinal results from the Massachusetts Male Aging Study. JCEM 2002; 87:589-98.


I've been in love with this band. This song rearranges me.
Trial run of my new camera and some new photographs. I am a sucker for industrial places. I really think the texture is beautiful. Here are a few shots I liked.
This ones for the good old days. These are my memories in red, blue, green.
..and my cover of one of my favorite songs by Radiohead.
Cliff jumping… And our trip to California.
GoPro camera's aren't just for jumping out of airplanes. This was filmed in Seattle. As you can see I'm a fan of time lapses.
I've been loving this old song for some reason this week. It seriously makes me feel 100 years old, coupled with my old man knees. I go back and forth on what version is my favorite (this or Nico's) but today its this one.
This still is my favorite documentary.
I love the way this story unfolds. It makes me wonder what, at the end of my life, will be my legacy? What will I leave and how exactly will it matter? At the end of our lives we have these wrinkles that tell our individual story. Now those wrinkles tell the story of their city.
Was just driving up to the film festival when I stumbled upon my future place of residence… It's a sign and who am I to argue with fate.
Conquered the snow storm of the year to get to the Sundance Film Festival last week. I barely arrived in one piece but once I did filmed a couple events for On The Green Carpet. Here is the first segment filmed at the Bing Lounge on Main Street.
This song is so cool. Then I found this video (not the official) synced with the clips Wax Tailor sampled in this song.
YES. Wes Anderson's films don't happen nearly enough. This one looks especially quirky, especially awesome, especially Wes Anderson. Bill Murray, Jason Schwartzman, hues of yellow, understated dialogue and toy box backdrops. I'm excited to see Frances McDormand, Bruce Willis and Ed Norton in this film as well. Not only am I looking forward to another Wes Anderson film but I'm looking forward to another Wes Anderson soundtrack.
Forget paper announcements, or if you're a traditionalist this totally enhances your announcements. For years I personally always wanted to do video announcements but never had thought of stop animation. I've seen lots of "Save the Date" video announcements but I like how simple and understated these are. Here are some cool ideas (that I didn't do) for "Save the Date" announcement videos.
As soon as I found out about HGTV's new dream home giveaway located in Midway I've been packing my bags and planning out my new life. Since this discovering I pretty much get up in the morning go straight to my computer and enter twice. Just investing in my future is all.
In my very reasonable and practical head I have convinced myself it's signed-sealed-and-delivered my new home. I've already considered who will live with me, what room will be the computer room, etc. I've considered stopping by and bringing the neighbors cookies. I've weighed the pro's and con's. Yes con's. For instance what a bummer it would be having to ALWAYS entertain all the people that would want to sit in my 8 person hot top. "Sorry! Hot tubs full. I wish I only had seven friends." I have already considered what grocery stores I'd have to shop at and been seriously stressed over who will take me to the airport. It's pretty tough when you're planning out your new life in your new dream home.
I've been working on my surprised face when they knock on the door.
If you're wondering where to find me in February I'll be here.
Only 9 days till the festival starts. Here's one film I'd like to see with Rashida Jones, Andy Samberg, and Elijah Wood.
I just came across this. I love it. I love the idea of exploring our personal visual sense. I love the idea that we always have five minutes in every day to do something. I love the idea of breaking up the ordinary to capture something extraordinary. I love the image of him and his friend in the car with pilot head sets on. I love this story.
I may or may not be a closet R. Kelly fan. Maybe some of you remember some of the ridiculous songs he came out with in the 90′s. Well if you do, and even if you don't, you should check this out because it's so cool.
A picture a day. That's one of my resolutions. With the abundance of cameras and devices I have at an arms reach from me at all times I'm surprised at how many undocumented gaps there are in my life. But hopefully at the end of 2012 I will have 365 photos. One for everyday. I won't post every picture. But here is one for January 5th.
This year was the first Christmas I got to enjoy with my whole family in a while. Here's the minute long summary.
I probably haven't put a puzzle together in 20 years. I can't imagine how my 8 yr old self could have put that together.
Some of you may be familiar with my first blog "Postcard from Mars," now deceased. Rest In Peace. But for those of you courageous and ready for change this is my new, grown up blog, now associated with my website ashleycabulagan.com which my Mother snidely remarked "Is so original." But, Mother, it is. With the help of google search I've pretty much confirmed I'm the only one. Ashley Cabulagan, that is. So while kicking off the New Year I thought I'd kick off the new blog, new site.. New life?
Here's to new beginnings… and a song I've been loving today that suits 2012′s fresh new start.
Here's some new stuff from my very talented and very funny friend Pablo, who currently resembles another friend of mine, Che Guevara. You should check out his stuff, if you want to.
This is his latest. You might recognize the awkward figure in the background attempting to throw in some harmonies. Cheers.
"Damn Everything But the Circus!"
Here you go Mom. This is my new favorite thing, walking on an inch wide line. It's really more fun than it sounds. I need a lot of practice but the goal is to someday do this a lot higher off the ground. I feel like maybe I missed my calling in life, the circus. I also feel like now that I'm practicing being a tight rope walker, if I were to trip walking down the sidewalk, it would be unacceptable.


Katharine & George Alexander Community Law Center has raised $ 32,145 from 25 gifts!
ARE YOU ALL IN FOR THE KATHARINE AND GEORGE ALEXANDER COMMUNITY LAW CENTER? YOU can help build momentum for the causes you care most about by saving the image below and sharing it on your social media sites to show your Bronco pride!
The Katharine & George Alexander Community Law Center (KGACLC) responds to the need for affordable legal services within Santa Clara County's most marginalized communities. Today, faced with the dramatic changes in our political, social and economic landscape, KGACLC renews its commitment to educating law students in the highest professional and ethical standards while serving the legal needs of the local low-income community.
By ensuring that the most vulnerable consumers are protected from unscrupulous business practices, that workers are paid for their work and treated equitably, and that immigrants are offered accurate legal advice and high-quality representation, we also ensure that our law students become effective attorneys and lifelong advocates for social justice.
GO ALL IN with the Katharine & George Alexander Community Law Center so that together we can continue to build unity through justice and education.


A disorientation guide looks to inform members of the Wellesley community about stories you wouldn't typically receive during orientation week. This guide was made to set the stage for a future Wellesley disorientation guide that includes more collaborations and statements from different orgs. This current tells the story of different student activism and tools Wellesley has to promote social good. ACCESS THE GUIDE HERE!!!!
ACCESS THE FILES HERE TO BEING COLLABORATING!!


This course is perfect for those who have a heavy schedule or are in full time employment and would struggle to commit to the same day weekly for group training courses.
The sessions are tailored to your individual needs and specific aspects of artistry that YOU wish to learn.
This course is suitable for anyone wanting to learn how to apply their own makeup professionally, or Qualified Artist's who wish to update their skills.
These classes must be taken between the hours of 10-6pm Monday-Friday.


I always save the paper towels that I use for clean-up when I'm painting. It makes me feel less wasteful knowing that these papers will be reused in future collages or in recycled paper making such as I'm going to show you today.
1. Tear up the paper towels and place about 1 c of shreds into a blender that's 1/2 to 3/4 full of water. Blend in short bursts and then blend until paper is pulped.
2. Place the hoop, concave side up, into the shallow dish and pour the pulp and water into it. Swirl and use your fingers to distribute the pulp over the organza.
3. Place the hoop onto stacked cloth towels and press with a sponge to drain out most of the water.
4. Add inclusions if using. You may need to add another smaller amount of pulp over top of these to hold them in place if using heavy objects. Soy silk or other roving fibers, small dried herbs, and colorants can be added without additional pulp.
5. Drizzle or spray some diluted matte medium over the top. Press gently to make sure the paper is saturated but the inclusions are not disturbed.
6. Turn hoop over onto a flat surface covered by waxed or freezer paper. Gently rub the organza to loosen the paper from the fabric. Loosen the hoop and remove, gently peeling off the organza.
7. Allow the paper to dry completely before using.
What a wonderful way for instant paper! Just think of all the things one could do with this! Its enough to get my fingers itching to start doing! Thank you for this mini tute.
Wow, I never knew you could do this! What a handy way to make use of all those paper towels I throw away.
What a fun way to upcycle used paper towels. Thank you, Cyndi.


An integral part of the Buying and Merchandising team, you will lead the Concession Business within Dover Street Market to drive trade and animate each of the concession spaces. You will be responsible for providing market expertise, establishing budgets, providing data analysis, forecasting and planning with ultimate responsibility for delivering seasonal and annual targets.
You will be hardworking and entrepreneurially spirited with exceptional organisation and communication skills that will help you to develop cross functional collaboration with the other departments within DSM and to build exceptional relationships with all the concessions brands.


If you've ever tried to use your phone or tablet in direct sunlight, you were probably frustrated by how difficult it is to see the screen. But instead of being frustrated, you may want to be thankful. Why? New research indicates that using a mobile device in the sunlight could actually be dangerous to your health.
When you use one of these devices in the sunlight, the screen is hard to see because it's reflecting so much light. But where exactly does that light end up? According to a small research study, much of it is actually hitting you in the face.
The researchers, Mary E. Logue and Dr. Barrett Zlotoff of the University of New Mexico were wondering if these screens worked a bit like tanning reflectors. To test this, they set up a mannequin head wearing a UVA/B light meter in a sunny area. In front of the mannequin, they placed a standard musician's sheet stand holding a variety of devices.
In one trial, they located the devices 16.5 inches from the UV sensor and angled them to resemble our typical posture when looking at these devices. They recorded UV exposure from 11 a.m. to noon and compared the readings obtained when using the devices to those obtained when the sheet stand was empty.
At first, they measured exposure simply using a regular magazine. That increased UV dosage exposure by 46%. That's high in and of itself. But then they found that an iPad2 and an 11-inch MacBook increased exposure by 85% and 75%, respectively.
For the second trial, the researchers positioned an iPhone5 12.25 inches from the sensor. Even this small device increased exposure by 36%.
While further research is needed to see if there's a direct link between using these devices outdoors and an increase in skin cancer risk, we know that over time UV exposure increases your chances of developing the disease and accelerates the aging of the skin. Logue explained, "The harmful effects of UVA and UVB rays have been well documented, and limiting exposure is the single most effective preventative measure an individual can take. Significant levels of UV exposure, such as those found in this study, increase cumulative lifetime UV dosage."
Logue continued, "While the best course of action is to limit smart device usage to indoors, this is obviously impractical for most people. We recommend covering the shoulders, wearing sunglasses, and wearing sunscreen, especially on the exposed areas of the neck and face."
I love my iPhone and iPad, so this was sobering news for me. It's a great reminder to always, always wear some kind of sunblock, even when I'm lazy and might think it's unnecessary, and to be more selective about using my devices outdoors. If the screen is difficult to make out, that's probably a sign that it's best to wait until later. Rather than straining your eyes and potentially damaging your skin, you can protect both by putting your devices away when you're in strong sunlight.


The Hatter (also called the Mad Hatter or Hatta) is a fictional character in Lewis Carroll's Alice's Adventures in Wonderland and its sequel Through the Looking-Glass. He is often referred to as the Mad Hatter, though this term was never used by Carroll. The phrase "mad as a hatter" pre-dates Carroll's works and the characters the Hatter and the March Hare are initially referred to as "both mad" by the Cheshire Cat, with both first appearing in Alice's Adventures in Wonderland, in the seventh chapter titled "A Mad Tea-Party".
The Hatter character, alongside all the other fictional beings, first appears in Lewis Carroll's 1865 novel Alice's Adventures in Wonderland. In it, the Hatter explains to Alice that he and the March Hare are always having tea because when he tried to sing for the Queen of Hearts, the foul-tempered monarch, at her celebration she sentenced him to death for "murdering the time", but he escapes decapitation. In retaliation, time (referred to as a "he" in the novel) halts himself in respect to the Hatter, keeping him and the March Hare stuck at 18:00 (or 6:00 pm) forever.
A hatter is a person engaged in hatmaking.
Luton Town F.C., team a.k.a. "The Hatters"
Stockport County F.C., team a.k.a. "The Hatters"
Plot: The Hatter tells his tale of a wonderful tea-party ruined by an annoying little girl named Alice.
Plot: Mark Watson used to lead a double life. One as a loving husband, and the other as a spy for the U.S. government. The past comes back to haunt Mark in the form of The Hatter; and international crime lord Mark had once spied on. It is here that The Hatter gives Mark the ultimate ultimatum. And Mark soon discovers that becoming a spy for The Hatter involves more than just pretending to be a fictional person... but actually becoming one.
Plot: New Zealand, 1860. Thirteen year old Scott Hunter becomes anxious about his father, a gold prospector, when his personal belongings return home without him. Scott goes off to Tuckers Gully in search and en route has various adventures, not least when he hitches a ride with a travelling showman, Marvello The Magician. On arriving at Tuckers Gully Scott finds that his father's colleague has been found murdered. Scott's father, whom no one has seen for some weeks, is the prime suspect and there is a price on his head. Scott finds that his father struck gold and is being held prisoner by the real murderers, Ryan and Dodd, and manages to convince the folk of Tuckers Gully the truth.
Stetson has already proven to be a bad team. So Duke easily dominated the Hatters.