intr goup th waner
Embed Size (px)
DESCRIPTIONThis book is about the theory of groups, it has a breve introduction to the groups.
INTRODUCTION TO GROUP THEORY
LECTURE NOTES BY STEFAN WANER
1. Complex Numbers: A Sketch 22. Sets, Equivalence Relations and Functions 53. Mathematical Induction and Properties of the Integers 124. Groups 155. Subgroups 196. The Permutation Groups 237. Cosets and Lagrange’s Theorem 278. Normal Subgroups and Quotient Groups 319. Homomorphisms 3410. Some Structure Theorems 4011. Group Actions 4212. The Sylow Theorems 48
Date: July 2003.
1. Complex Numbers: A Sketch 2
1. Complex Numbers: A Sketch
A complex number is just a pair, z = (a, b) of real numbers. We usuallywrite this pair in the form z = a + ib, where the “+” and “i” are justdecorations (for now). The number a is called the real part of z, while b iscalled the imaginary part of z. We denote the set of all complex numbersby C. Note that we can represent any complex number z = a+ ib ∈ C by apoint in the plane.
Examples 1.1. In class of complex numbers and their locations in the plane
Definition 1.2. We define addition and multiplication of complex numbersas follows:
(1) (a+ ib) + (c+ id) = (a+ c) + i(b+ d)(2) (a+ ib)(c+ id) = (ac− bd) + i(ad+ bc)
In other words, we add complex numbers by adding their real and imaginaryparts, and multiply them by treating i as a square root of −1.
Notation 1.3. We use the following shorthand notation:a+ i · 0 = a (That is, (a, 0) = a)0 + ib = ib (That is, (0, b) = ib)0 + i1 = i (That is, (0, 1) = i)
Then we see that the sum of a and ib is indeed the single complex numbera+ ib. In other words, we can now think of a+ ib as a sum rather than asa wild and crazy way of writing (a, b).
(1) i2 = −1 (Check it and see; remember that i is just shorthand for thecomplex number +i1)
(2) For every complex number z, we have 1 · z = z · 1 = z(3) Addition and multiplication of complex numbers obey the same rules
(commutativity, associativity, distributive laws, additive identity,multiplicative identity) as the real numbers. We’ll see in a minutethat there are also inverses.
Examples 1.5. Illustrating the geometry of addition and multiplication:In class.
Definition 1.6. The magnitude of the complex number z = a+ib is givenby the formula
|z| =√a2 + b2
(This is just its distance from the origin) Also, we define:
z = a− ib,
called the complex conjugate of z.
1. Complex Numbers: A Sketch 3
Examples in class
Now we notice that:zz = |z|2
In other words:z · z
But this says that z/|z|2 is the multiplicative inverse of z. In other words,
|z|2Examples in class
We now look at the polar form, and we can write
z = r cos θ + ir sin θ = r(cos θ + i sin θ)
We also write this as reiθ. (Explanation in class)
Notes 1.7.(1) If z = reiθ, then r = |z|.(2) the identity [r(cos θ + i sin θ)][s(cosφ + i sinφ)] = rs(cos(θ + φ) +
i sin(θ + φ) translates to reiθ + reiφ = rei(θ+φ), which is what weexpect from the laws of exponents.
(3) Addition and multiplication of complex numbers. Also, this gives usthe key to the geometric meaning of multiplication.
(4) If we multiply a complex number by itself repeatedly, we now get:
[r(cos θ + i sin θ)]n = rn(cosnθ + i sinnθ),
which is known as De Moivres formula. We can use it to find nthroots of any complex number: take the nth root of the magnitude,and divide the angle θ by n. We can also divide θ + 2π by n to getanother nth root, and θ+4π, θ+6π, etc. They start repeating whenwe get to θ + 2nπ. In other words:
There are n− 1 different nth roots of any complex number.
Examples 1.8.A. Find all the 4th roots of i.B. Find all the 5th roots of 32eiπ/3.C. Find all the roots of the equation z3 = i.
If we choose r = 1 in De Moivres formula, this places us on the unit circle,and we find all kinds of nth roots of 1.
Definition 1.9. The primitive nth root of unity is the complex numberω = e2πi/n. Note that all the other nth roots of unity are powers of ω. Inother words, the n nth roots of unity are:
1 = ω0, ω, ω2, . . . , ωn−1.
1. Complex Numbers: A Sketch 4
Note This gives us another (easy) way of getting all the nth roots of anycomplex number: Find one of them, and then multiply it by the differentnth roots of unity above!
Exercise Set 1.1. Find all the 8th roots of unity, sketch their position on the unit
circle, and represent them both in Cartesian form (x+ iy) and polarform (reiθ).Which of them are real, and which pure imaginary?
2. Find all the 5th roots of 32eiπ/3
3. (a) Show that, if z is a complex number, then zn = zn for allintegers n (including negative ones).
(b) Show that, if ω is an nth root of unity, then so are ω and ωr forevery integer r.
(c) Show that, if ω is an nth root of unity, then ω−1 = ωn−1.4. (a) The set of Gaussian integers is defined as the set Z[i] of all
complex numbers of the form m0 +m1i, where m0 and m1 areintegers. Prove that products of Gaussian integers are Gaussianintegers.
(b) If ω is an nth root of unity, define Z[ω], the set of generalizedGaussian integers to be the set of all complex numbers of theform m0 +m1ω +m2ω
2 + · · ·+mn−1ωn−1,where n and mi are
integers. Prove that products of generalized Gaussian integersare generalized Gaussian integers.
2. Sets, Equivalence Relations and Functions 5
2. Sets, Equivalence Relations and Functions
A set is an undefined “primitive” notion. Intuitively, it refers to a col-lection of things called elements. If a is an element of the set S, we writea ∈ S. If a is not an element of the set S, we write a /∈ S. Some importantsets are:
Z, the set of all integersN, the set of all natural numbers (including 0)Z+, the set of all positive inegersR, the set of all rational numbersR, the set of all real numbersC, the set of all complex numbers
We can describe a set in several ways:
(1) by listing its elements; e.g.. S = 6, 66, 666(2) in the form x | P (x), where P (x) is a predicate in x, for instance
S = x | x is a real number other than 6
T = x ∈ Z | x odd
Note: Two sets are equal if they have the same elements. That is,A = B means x ∈ A⇔ x ∈ B
Definitions 2.1. Let A and B be sets.We say that A is a subset of B and write A ⊂ B if x ∈ A⇒ x ∈ B.
A ∩B is the intersection of A and B, and is given byA ∩B = x | x ∈ A and x ∈ B
A ∪B is the union of A and B, and is given byA ∪B = x | x ∈ A or x ∈ B
∅ is the empty set; ∅ = x | F (x), where F (x) is any false predicatein x, such as “x = 3 and x 6= 3” or “Your math instructor drives ared mustang with x doors.”
A−B is the complement of B in A, and is given byA−B = x | x ∈ A and x /∈ B
A×B is the set of all ordered pairs,A×B = (a, b) | a ∈ A, b ∈ B
If Aα | α ∈ Ω is any collection of sets indexed by Ω, then:⋂α∈ΩAα is the intersection of the Aα and is given by⋂
α∈ΩAα = x | x ∈ Aα for all α ∈ Ω
2. Sets, Equivalence Relations and Functions 6⋃α∈ΩAα is the union of the Aα and is given by⋃
α∈ΩAα = x | x ∈ Aα for some1α ∈ Ω
Note: To prove that two sets A and B are equal, we need only prove thatx ∈ A⇔ x ∈ B. In other words, we must prove two things:
(a) A ⊂ B (ie., x ∈ A⇒ x ∈ B)(b) B ⊂ A (ie., x ∈ B ⇒ x ∈ A)
Lemma 2.2. The following hold for any three sets A, B and C and anyindexed collection of sets Bα (α ∈ Ω):
(1) Associativity:A ∪ (B ∪ C) = (A ∪B) ∪ C A ∩ (B ∩ C) = (A ∩B) ∩ C
(2) Commutativity:A ∪B = B ∪A A ∩B = B ∩A
(3) Identity and IdempotentA ∪A = A, A ∩A = AA ∪ ∅ = A, A ∩ ∅ = ∅
(4) De Morgan’s Laws:A−(B∪C) = (A−B)∩(A−C), A−(B∩C) = (A−B)∪(A−C)
(5) Distributive Laws:A∪(B∩C) = (A∪B)∩(A∪C), A∩(B∪C) = (A∩B)∪(A∩C)
Proof We prove (1), (2), (3) and a bit of (4) in class. The rest you willprove in the exercise set.
Definition 2.3. A partitioning of a set S is a representation of S as aunion of disjoint subsets Sα, called partitions:
where Sα ∩ Sβ = ∅ if α 6= β.
Examples 2.4.A. The set Z can be partitioned into the odd and even integers.B. Z = 3Z ∪ (1 + 3Z) ∪ (2 + 3Z), where m + 3Z = m + 3n | n ∈ Z.
(You will spell this out in more detail in the exercise set.)C. The set of n × n matrices can be partitioned into subsets each of
which contains matrices with the same determinant.
Definition 2.5. A relation on a set S is a subset R of S×S. If (a, b) ∈ R,we write aR b, and say that a stands in the relation R to b.
1That is, at least one
2. Sets, Equivalence Relations and Functions 7
Examples 2.6.A. Equality on any set AB. 6= on any set AC. < on ZD. m ≈ n if m− n ∈ 3Z, on ZE. Row equivalence on the set of m× n matricesF. Any partitioning of a set S gives one: Define aR b if a and b are in
the same partition.
Definition 2.7. An equivalence relation on a set S is a relation ≈ on Ssuch that, for all a, b and c ∈ S:
(1) a ≈ a (Reflexivity)(2) a ≈ b⇒ b ≈ a (Symmetry)(3) (a ≈ b and b ≈ c) ⇒ a ≈ c (Transitivity)
Examples 2.8.A. Equality on any setB. Equivalence mod k on Z (in class)C. Row equivalence on the set of m× n matricesD. Any partition on S yields an equivalence relation
Definition 2.9. If ≈ is an equivalence relation on S, then the equivalenceclass of the element s ∈ S is the subset
[s] = t ∈ S | t ≈ s
Lemma 2.10. Let ≈ be any equivalence relation on S. Then(a) If s, t ∈ S, then [s] = [t] iff s ≈ t.(b) Any two equivalence classes are either disjoint or equal.(c) The equivalence classes form a partitioning of the set S.
Theorem 2.11. Equivalence Relations “are” PartitionsThere is a one-to-one correspondence between equivalence relations on aset S and partitions of S. Under this correspondence, an equivalence classcorresponds to a set in the partition.
Examples 2.12.A. Z/nZ is the set of equivalence classes of integers mod n. [r] = [s] iff
r − s ∈ nZ. We look at these equivalence classes explicitly in class.B. Construction of the rationals Define a relation on Z × Z∗ by
(m,n) ∼ (k, l) iff ml = nk. The quotient mn is defined to be the
equivalence class of (m,n). (Exercise set)
Definition 2.13. Let A and B be sets. A map or function f : A→ B is atriple (A,B, f) where f is a subset of A×B such that for every a ∈ A, thereexists a unique b ∈ B (that is, one and only one b ∈ B) with (a, b) ∈ f . Werefer to this element b as f(a). A is called the domain or source of f andB is called the codomain or target of f .
2. Sets, Equivalence Relations and Functions 8
Notes:(1) We think of f a rule which assigns to every element of A a unique
element f(a) of B, and we can picture a function f : A→ B as shownin the figure.
(2) The codomain of f is not the “range” of f ; that is, not every elementof B need be of the form f(a).
(3) The sets A and B are part of the information of f . For instance,specifying f by saying only “f(x) = 2x− 1” is not sufficient becausewe have not specified the domain and codomain. We should insteadsay something like this:
“Define f : R → R by f(x) = 2x− 1.”
Examples 2.14. Some in class, plus:A. If A is any set, we have the identity map 1A : A→ A; 1A(a) = a for
every a ∈ A.B. If B ⊂ A, then we have the inclusion map ι : B → A; ι(b) = b for all
b ∈ BC. The empty map ∅ : ∅ → A for any set A.
Definition 2.15. Let f : A → B be a map. Then f is injective (or one-to-one) if
f(x) = f(y) ⇒ x = y
In other words, if x 6= y, then f(x) cannot equal f(y).
Examples 2.16.A. f : R → R; f(x) = 2x− 1 is injective.B. f : R → R; f(x) = x2 + 1 is not.C. The identity 1A : A→ A is injective for every set A.D. The inclusion ι : B → A is injective for every set A and every subset
B ⊂ A.
Definitions 2.17. Let f : A → B be a map, and let C ⊂ A. Then theimage of C under f is the subset
2. Sets, Equivalence Relations and Functions 9
f(C) = f(c) | c ∈ C(See the figure.)
The image of f is defined asIm f = f(A).
f is surjective (or onto) if Im f = B. In other words,
b ∈ B ⇒ ∃a ∈ A such that f(a) = b
Thus, f “hits” every element in B (see figure).
f is surjective iff Im f = B
Note: If f : A→ B, then f(A) is sometimes called the range of f .
Examples 2.18.A. f : R → R; f(x) = x2 + 1. Find f(R) and f [0,+∞).B. Identity maps are always surjective.C. The inclusion ι : C → B is surjective iff C = B.
2. Sets, Equivalence Relations and Functions 10
D. The canonical projections of a (possibly infinite) product.E. Let S be any set and let ≈ be an equivalence relation on S. Denote
the set of equivalence classes in S by S/≈. Then there is a naturalsurjection ν : S → S/≈.
Lemma 2.19. Let f : A→ B. Then:(a) f−1(f(C)) ⊃ C for all C ⊂ A with equality iff f is injective;(b) f(f−1(D)) ⊂ D for all D ⊂ B with equality iff f is surjective;
Proof Exercise Set 2. We’ll prove (a) in class.
Definition 2.20. f : A → B is bijective if it is both injective and surjec-tive.
Examples 2.21.A. Exponential map R → R+
B. Square root functionC. Inverse Trig functionsD. Multiplication by a non-zero real numberE. The identity map on any set
Definition 2.22. If f : A → B and g : B → C, then their composite,g f : A→ C, is the function specified by g f(a) = g(f(a)).
g f(Example in class.)
Lemma 2.23. Let f : A→ B and g : B → C. Then:(a) If f and g are injective, then so is g f .(b) If f and g are surjective, then so is g f .(c) If g f is injective, then so is f .(d) If g f is surjective, then so is g.
Definition 2.24. f : A → B and g : B → A are called inverse maps ifg f = 1A and f g = 1B. In this event, we write g = f−1 (and say thatg is the inverse of f) and f = g−1. If f has an inverse, we say that f isinvertible.
Theorem 2.25 (Inverse of a Function).(a) f : A→ B is invertible iff f is bijective(b) The inverse of an invertible map is unique.
We prove (a) in class and leave (b) as an exercise.
Exercise Set 2.1. Prove Lemma 2.2 ( 4) and ( 5).
2. Sets, Equivalence Relations and Functions 11
2. Prove thatA×
3. (cf. Example 2.4[B]) Prove that Z can be partitioned as Z = 3Z ∪(1 + 3Z) ∪ (2 + 3Z), where m + 3Z = m + 3n | n ∈ Z. (Youmust show that Z is indeed the union of the three sets shown. [Hint:consider the remainder when an arbitrary integer is divided by 3.]
4. Give an example of a relation on Z which is:a. reflexive and symmetric but not transitive;b. transitive and reflexive but not symmetric;c. transitive and symmetric but not reflexive.
5. Verify that the relation on Z×Z∗ given by (m,n) ≈ (k, l) iff ml = nkis an equivalence relation.
6. a. Let M(n) be the set of n × n matrices, let P be some fixedn× n matrix, and define f : M(n) → M(n) by f(A) = PA. (fis called “left translation by P.”) Show that f is injective iff Pis invertible.
b. Let f be as in (a). Show that f is surjective iff P is invertible.7. Prove Lemma 2.19.8. a. Give an example of a map f : A → B and a map g : B → C
with g f injective but g not injective. (See Lemma 2.23.)b. Give an example of a map f : A → B and a map g : B → C
with g f surjective but f not surjective. (See Lemma 2.23.)9. a. By citing appropriate results in these notes, give a two-line proof
that if f : A→ B and g : B → C are bijective, then so is g f .b. Give an example of a map f : A → B and a map g : B → C
with g f bijective, but with neither f nor g bijective.10. Prove Theorem 2.25(b).11. Prove that composition of functions is associative: (f g) h =
f (g h) and unital: f 1A = 1B f = f for all f : A→ B.12. Let f : A→ B, and define a relation on A by a ≈ a if f(a) = f(a′).
Show that ≈ is an equivalence relation on A.
3. Mathematical Induction and Properties of the Integers 12
3. Mathematical Induction and Properties of the Integers
The Axiom of Mathematical Induction is one of the central axioms ofarithmetic. Here is one of the forms it can take:
Axiom of Mathematical InductionIf S is any subset of N such that:
(a) 0 ∈ S;(b) n ∈ S ⇒ n+ 1 ∈ S,
then S = N.
The following is a theorem in “meta-mathematics”:
Theorem 3.1 (Principle of Mathematical Induction).If P (n) is any proposition about the natural number n such that:
(a) P (0) is true;(b) If P (n) is true, then P (n+ 1) is true,
then P(n) is true for all n ∈ N.
Proof: Let S = n ∈ N | P (n) is true . . .
Corollary 3.2 (General Principle of Mathematical Induction).If P (n) is any proposition about the natural number n ≥ k such that:
(a) P (k) is true;(b) If P (n) is true, then P (n+ 1) is true,
then P(n) is true for all n ≥ k.
Proof: Let Q(n) be the proposition “P (n + k) is true,” and apply thetheorem to Q.
Examples 3.3. We prove the following by induction:A. 1 + 2 + · · ·+ n = n(n+ 1)2 for all n ≥ 1.B. De Morgan’s Law for finite unions: If A, Bi (i = 1, 2, . . . ) are any
C. Every polynomial over Z factors as a product of irreducible polyno-mials over Z.
Definition 3.4. The integer a divides the integer b if there exists an integerk such that b = ak. When this is the case, we write a|b.Definition 3.5. The positive integer h is the greatest common divisor(gcd) or highest common factor (hcf) of a and b if
(a) h is a divisor of a and b; (we write h|a and h|b)(b) If d is a divisor of a and b, then d is a divisor of h. That is,
d|a and d|b⇒ d|h
3. Mathematical Induction and Properties of the Integers 13
We denote the hcf h of a and b by (a, b).
Note: The hcf is always positive, so that (±a,±b) = (a, b).
Proposition 3.6 (Existence and Properties of the hcf).If a and b are integers, not both 0, then (a, b) exists. Moreover, there existintegers m and n such that
ma+ nb = (a, b)
Proof: Here is the very elegant proof in Herstein: Let
M = ma+ nb | m,n ∈ ZThen M certainly contains positive integers (since a and b are not bothzero), so let h be the smallest positive element of M . We claim that h is thehcf, which not only proves existence, but also the propertyh = ma+ nb.
Indeed, we need to show properties (a) and (b) above.
Property (a): Let us prove that h is a divisor of a. If a = 0, then h iscertainly a divisor of a. If a 6= 0, then by the division algorithm we candivide a by h to obtain
a = ph+ r with 0 = r < h
(that’s even true is a is negative!) Hence
r = a− ph
= a− p(ma+ nb)
= (1− pm)a+ (−pn)b ∈M.
But now r ∈ M is non-negative, and is smaller than the smallest positiveelement, which forces it to be zero. In other words, h is a divisor of a, asclaimed.
Property (b): If d is a common divisor of a and b, we must show that d isalso a divisor of h. But the hypothesis implies that
a = kd and b = ld
for some k and l. Hence
h = ma+ nb
= m(kd) +m(ld) = d(mk +ml),
showing that h is divisible by d, as required.
Definition 3.7. The integers a and b are relatively prime if (a, b) = 1.
Corollary 3.8 (Relatively Prime Integers).If a and b are relatively prime, then there exist integers m and n such that
ma+ nb = 1.
3. Mathematical Induction and Properties of the Integers 14
Example in class
Definition 3.9. The integer p > 1 is prime if its only divisors are ±1 and± p. It follows that p is prime iff (n, p) = 1 or p for every integer n.
Lemma 3.10.If (a, b) = 1 and a | bc, then a | c.
Proof: Choose m and n such that ma+ nb = 1. Multiplying by c gives
mac+ nbc = c.
Since a divides both terms on the left (the second term by hypothesis), itmust divide the term on the right.
Corollary 3.11.(a) If p is prime, and p | bc, then p | b or p | c. More generally,(b) If p is prime, and p divides any product of integers, then it divides
at least one of them.
Proof: For part (a), assume p | bc but p - b. Then (p, b) = 1, so by thelemma, p|c. Part (b) follows by an inductive argument. Extra Reading: Unique factorization of integers into primes
Exercise Set 3.Prove the the statements in Exercises 1 to 3 by induction:
1. 12 + 22 + · · ·+ n2 =n(n+ 1)(2n+ 1)
6for n ≥ 1
2. 13 + 23 + · · ·+ n3 =n2(n+ 1)2
4for n ≥ 1
3. 1 + 3 + 5 + · · ·+ (2n− 1) = n2 for n ≥ 1
4. LetR =[cos θ − sin θsin θ cos θ
]. Prove by induction thatRn =
[cosnθ − sinnθsinnθ cosnθ
]for all n ≥ 1.
The following are taken from Herstein’s Topics in Algebra (p. 23):5. Prove: If a | b and b | a, then a = ±b.6. Prove: If a | x and b | x, and (a, b) = 1, then ab | x.7. (A step-by-step approach to Herstein’s starred problem) Let p be a
prime number.(a) Use the binomial theorem to show that (n + 1)p − (np + 1) is
divisible by p for all n ≥ 1.(b) Use part (a) and induction to prove that, for all positive integers
a, ap − a is divisible by p.(c) Deduce that (b) is true for all integers a.
Note: You have proved that ap− a is divisible by p for every primep. That is, ap − a ≡ 0 mod p, as stated in Herstein’s problem.
4. Groups 15
Definition 4.1. A binary operation on a set S is a map ∗ : S × S → S.In other words, the operation ∗ assigns to each pair (s, t) of elements in San element ∗(s, t), which we shall write as s ∗ t, of S.
Examples 4.2.A. Addition, subtraction and multiplication on RB. Division and multiplication in Q∗C. Composition in Map(X,X), the set of maps X → XD. Multiplication of n× n matricesE. Concatenation of strings in a given set of symbolsF. N is not closed under subtraction; hence subtraction is not a binary
operation on N.G. The quotient a/b is not defined for every pair of real numbers (a, b),
Hence division is not a binary operation on R.
Recall that Z/nZ is the set of equivalence classes of integers modulo n.
Lemma 4.3 (Addition modulo n).The operation +: Z/nZ× Z/nZ → Z/nZ given by [m] + [n] = [m+ n] is awell defined binary operation.
Some of these operations have the nice properties we ascribe to a group:
Definition 4.4. A group (G, ∗) is a set G together with a binary operation∗ on G such that:
(a) ∗ is associative.(b) (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ G.(c) There is an element e ∈ G called an identity element such that
e ∗ g = g ∗ e = g for every g ∈ G.(d) For every g ∈ G, there exists an element g′ ∈ G, called an inverse
of g such that g ∗ g′ = g′ ∗ g = e.
Examples 4.5.A. (Z,+)B. (Q∗,×)C. (M(m,n),+)D. (GL(n; Q),×), (GL(n; R),×), and (GL(n; C),×)E. The set
Cn = ω0, ω, ω2, . . . , ωn−1 ⊂ Cof nth roots of unity, under multiplication.
F. (Z/nZ,+) Compare its group structure with that of Cn.G. The set of all invertible maps R → R under compositionH. The set SA of all bijections of a set A, under compositionI. The unit circle,
S1 = z ∈ C | |z| = 1 under multiplication
4. Groups 16
J. Z is not a group under subtraction.K. Z is not a group under multiplication.L. Q is not a group under multiplication
Note: When we do not want to be explicit about the group operation ∗,we shall leave it out, and write a ∗ b as ab or sometimes as a + b whenappropriate. Similarly, we shall write a group as G rather than (G, ∗) whenthe group operation is understood.
Lemma 4.6 (Cancellation Law, Uniqueness of Identity and Inverses).If G is a group, then the following are true:
(a) Left Cancellation: If g, h and k have the property that gh = gk, thenh = k.
(b) Right Cancellation: If g, h and k have the property that hg = kg,then h = k.
(c) Uniqueness of the Identity: If e and e′ are both identities in G, thene = e′.
(d) Uniqueness of Inverses: If g′ and g′′ are both inverses of g ∈ G, theng′ = g′′.
Note: As a consequence of part (d) of the lemma, we shall speak of theinverse of g and write it as g−1. Similarly, we shall speak of the identityelement of a group.
Lemma 4.7 (Product of Inverses). If G is a group and a, b ∈ G, then(ab)−1 = b−1a−1.
The proof is in the exercise set.
Definition 4.8. An abelian group is a groupG satisfying the commutativelaw:
ab = ba for all a, b ∈ G.
Examples 4.9. Spot which of the above examples of groups are abelian.
Further Important Examples of Groups 4.10.A. The Dihedral groups Dn
Dn is the set of symmetries of the regular n-gon (n rotations and nreflections). Illustrated in class. If a is rotation through 2π/n, andif b is reflection in the x-axis, then we can write Dn as:
Dn = e, a, a2, . . . , an−1, b, ba, ba2, . . . , ban−1 Checking that this indeed gives us a group is tedious, but we shallsee in the exercises how to avoid this by realizing Dn as a group of2×2 matrices. We can multiply elements according to the rule: ba =a−1b. In terms of “generators and relations” we can also describeDn as:
Dn = 〈 a, b | an = b2 = e, bab−1 = a−1 〉
4. Groups 17
B. The Symmetric groups Sn
Sn is defined to be the group of permutations (self-bijections) of theset 1, 2, 3, . . . , n . In other words, Sn = S 1,2,3,...,n . We look atthe examples S2 and S3 and their multiplicative structure in class.
C. The Quaternion group Q8
Q8 is defined by
[1 00 1
[−1 00 −1
[0 1−1 0
[0 −11 0
[0 ii 0
[0 −i−i 0
[i 00 −i
[−i 00 i
]The group operation is matrix multiplication, and i is the familiarcomplex number. We abbreviate these elements as follows:
Q8 = 1,−1, i,−i, j,−j, k,−k
In class, we verify the following:
i2 = j2 = k2 = −1, and ij = k = −ji, jk = i = −kj, ki = j = −ik
Exercise Set 4.
1. List all six elements of D3, together with its multiplication table.2. Show that,if G is a group and g and g′ are such that gg′ = e, then
g′ = g−1.Hand In:
1. Show that the operation · : Z/nZ×Z/nZ → Z/nZ given by [m]·[k] =[mk] is a well defined binary operation.
2. (a) Prove Lemma 4.7.(b) Show that, if G is abelian, then (ab)−1 = a−1b−1 for all a, b ∈ G.(c) Give an example to show that (ab)−1 6= a−1b−1 in general.
3. Show that, if G is a group and g ∈ G is such that gn = gm for somem 6= n, then there exists an integer r with gr = e.
4. Show that, if G is a finite group and g ∈ G, then there exists apositive integer r with gr = e.
5. Prove that, in any finite group G, the inverse of each element is apower of itself.
6. Recall from the exercises on induction that if R =[cos θ − sin θsin θ cos θ
then one has Rn =[cosnθ − sinnθsinnθ cosnθ
]for all n ≥ 1.
(a) Let θ = 2π/n . Show that Rn = I, the identity matrix, and that I,R,R2, . . . , Rn−1 is a group under matrix multiplication.Which group does it remind you of?
(b) Let R and θ be as above, and let T =[1 00 −1
]. Show that
T 2 = I, and that TRT−1 = R−1. Deduce that
4. Groups 18
I,R,R2, . . . , Rn−1 T, TR, TR2, . . . , TRn−1is a realization of Dn, and use this to construct the multiplica-tion table for D3.
7. Product Groups If G and G′ are groups, define an operation onG×G′ by (a, a′)(b, b′) = (ab, a′b′). Show that this turns G×G′ intoa group.
5. Subgroups 19
First, some comments and definitions.
Remark The notation + for the group operation will only be used forabelian groups, and then only for groups in which it stands for what iscommonly thought of as the sum. Such a group will be called additive andits identity element will be written as 0 rather than e.
Definitions 5.1.(1) The order of the finite group G is the cardinality of G as a set, and
will be denoted by |G|.(2) If S is a subset of G such that for every s and t ∈ S, one has st ∈ S,
then we shall say that S is closed under the group operation inG. Note that S then inherits a binary operation from G. If S is asubset of G which is closed under the group operation, we refer tothe resulting binary operation on S as the induced operation onS .
Examples 5.2.A. 2Z ⊂ Z is closed under the group operation (+) of Z.B. N ⊂ Z is also closed under addition.C. The set of odd numbers is not a closed subset of Z under addition.
Definition 5.3. A subgroup of G is a subset H ⊂ G such that:(a) H is closed under the group operation;(b) H is a group in its own right under the induced operation. If H is a
subgroup of G, we shall write H < G.
Examples 5.4.A. Every group is a subgroup of itself.B. The subset e ⊂ G is a subgroup.C. 2Z ⊂ Z is a subgroup because it is a group in its own right.
Note: A subgroup H of G must be a non-empty subset of G, since being agroup in its own right implies that it contains the identity.
In practice, the following criterion is extremely useful in checking that agiven subset of G is a subgroup:
Proposition 5.5 (Test for a Subgroup). A subset H ⊂ G is a subgroup ofG iff:
(a) H is non empty and closed under the group operation(b) H is closed under inverses: if h ∈ H, then h−1 ∈ H.
In words, this tells us that for a subset to be a subgroup, it must be nonemptyand contain the products and inverses of all its elements. It also gives usthe following mechanical test:
5. Subgroups 20
How to check that H is a subgroup of G(a) Show that H 6= ∅, and if a and b ∈ H, then ab ∈ H(b) Show that, if h ∈ H, then h−1 ∈ H.
We can now give many more examples of subgroups.
Examples 5.6.A. The subgroup SL(n) of GL(n) consisting of n × n matrices with
determinant +1B. G = GL(n); H the set of upper triangular matricesC. Find all the subgroups of C6.D. Find all the subgroups of D2.E. Some finite subgroups of the circle groupF. Subgroups of ZG. The continuous functions in (RR,+)
Definition 5.7. A group G is called cyclic if it contains an element g suchthat every element of G is a (possibly negative) power of g. We refer to gas a generator of G.
Examples 5.8.A. Generators of Cn; for g = ωr to be a generator, some power of it
must be ω. This translates to an interesting requirement on r.B. Generators of ZC. Generators of mZD. Dn is not cyclic.E. Q is not cyclic under ×
Definition 5.9. A proper subgroup of G is a subgroup H < G such thatH 6= G.
Definition and Proposition 5.10. If g ∈ G, let 〈g〉 denote the subset gn | n ∈ Z . Then 〈g〉 is a subgroup of G, called the cyclic subgroupgenerated by g.
Examples 5.11.A. Look at various such subgroups of Cn
B. 〈a〉 and 〈b〉 are subgroups of Dn.C. The subgroup of GL(n) generated by a rotation matrixD. Cyclic subgroups of Z
Theorem 5.12 (Subgroups of Cyclic Groups).Every subgroup of a cyclic group is cyclic.
Proof: Let G be a cyclic group, so that G = 〈g〉, say, and let H < G. ThenH is a set of powers of g. Choose n to be the smallest positive exponent ofelements in H;
5. Subgroups 21
n = min i ∈ N | i > 0 and gi ∈ H .Then I claim that every element of H is a power of a = gn, giving the result.Indeed, if h ∈ H is not the identity, then either h or h−1 is of the form gm
with m > 0, so that m = n. Dividing m by n givesm = qn+ r
with r < n or r = 0, whencegm = (gn)qgr, givinggr = gm(g−n)q,
a product of elements of H, showing that gr ∈ H. By the choice of m, wemust have r = 0, givingh (or h−1) = gm = (gn)q,
proving the result.
Corollary 5.13 (Classification of subgroups of Z).Every subgroup of Z is cyclic and of the form nZ.
Exercise Set 5.
1. Which of the following are closed under the group operation?(a) The set of upper-triangular n×nmatrices as a subset of (M(n, n),+)(b) The set of upper-triangular n×nmatrices as a subset of (GL(n,R),×)(c) The set of pure rotations in Dn
(d) The set of transpositions in Sn
2. Which of the following are subgroups of (C,+)?(a) R(b) 7Z(c) iR (the set of pure imaginary numbers including 0)(d) 3Q
Hand In:1. Which if the following are subgroups of GL(n,R)? Justify your
answer in in each case.(a) n× n matrices with determinant 2(b) n× n matrices with determinant 1(c) n× n matrices with determinant a power of 3(d) n× n matrices with determinant ±1(e) n× n matrices A with AAT = I
2. (a) Prove that, if H and K are subgroups of G, then H ∩K is alsoa subgroup of G.
(b) Show that the intersection of any collection of subgroups of agroup G is a subgroup of G.
(c) True or false? The union of any two subgroups of a group isagain a subgroup. Prove or give a counterexample.
3. One-Step Test for SubgroupsProve that a subset H ⊂ G is a subgroup of G iff H is nonempty,and hk−1 ∈ H whenever h and k ∈ H.
5. Subgroups 22
4. Describe the cyclic subgroups of GL(2,R) generated by:
(a)[1 10 1
[0 −1−1 0
]5. Show that a cyclic group with only one generator has order at most
two. [Hint: consider the inverse of a generator . . . ]6. Prove that every cyclic group is abelian.7. (a) Show that any group G with no proper subgroups other than
e is cyclic.(b) Now deduce that, in fact, such a group must be finite of prime
order.8. Order of an Element of G
If G is a group and g ∈ G, the order of g is the smallest positiveinteger k with the property that gk = e. If no such k exists, we saythat g has infinite order.(a) Find the order of ω3 in C666
(b) If g ∈ G, show that g has order k iff |〈g〉| = k9. Conjugate Subgroups
Let H < G be a subgroup, and let g ∈ G. Define:gHg−1 = ghg−1 | h ∈ H
Show that gHg−1 is a subgroup of G. We call this subgroup theconjugate of H under g.
6. The Permutation Groups 23
6. The Permutation Groups
Recall that Sn is the group of all permutations of the set 1, 2, . . . , n .
Definition 6.1. The cycle (n1, n2, . . . , nm) of distinct integers ni ≤ n isthe element σ of Sn defined by:
ni+1 if j = ni and i ≤ m− 1n1 if j = nm
j otherwiseWe call m the length of the cycle.
Examples 6.2. In class; including 1-cycles, 2-cycles, etc.
Notes(a) (n1, n2, . . . , nm) = (nm, n1, n2, . . . , nm−1) = . . .(b) (n1, n2, . . . , nm)−1 = (nm, nm−1, . . . , n1)
Multiplying Cycles Practice examples in class, and specifically (1, 2, 3, 4)(4, 5, 6, 7)
Examples 6.3. Some Permutation GroupsA. S3
C. Dn for various n may be represented as a group of permutations—namely, what the rotations and reflections do to the vertices
D. Rigid motions of the cube
Definition 6.4. The two cycles (n1, n2, . . . , nm) and (r1, r2, . . . , rs) are dis-joint if n1, n2, . . . , nm and r1, r2, . . . , rs are disjoint as sets.
Note: Multiplication of disjoint cycles is commutative.
If σ ∈ Sn, define an equivalence relation on 1, 2, . . . , n by a ≈ b if thereexists an r ≥ 0 with σr(a) = b.
Lemma 6.5. The relation defined above is indeed an equivalence relation.(Proof in the Exercise set)
Definition 6.6. The equivalence classes of the relation defined above arecalled the orbits in 1, 2, . . . , n under σ.
Lemma 6.7. Each orbit in 1, 2, . . . , n under σ ∈ Sn has the form
k, σ(k), σ2(k), . . . , σj−1(k) for some k ∈ 1, 2, . . . , n .
Proof: Let O be any orbit under σ, and choose k ∈ O. Consider the subsetS = k, σ(k), σ2(k), . . . ⊂ 1, 2, . . . , n . By definition of the equivalencerelation, every element of the orbit containing k must be a power of σ appliedto k, and hence in S. Conversely, every element of S must be in the same
6. The Permutation Groups 24
orbit as k. It follows that the orbit of O of k is exactly the set S. That is,O = S.
What, exactly, does S look like? Since S cannot have more than n ele-ments, we must have repetitions; that is, σi(k) = σj(k) for some i < j < n.Let j be the smallest integer ≥ 1 such that σi(k) = σj(k) for some i < j.Claim: i = 0; that is, σj(k) = 1. Indeed, if i > 0, (so that j ≥ 2) then applyσ−1 to both sides of the equation σi(k) = σj(k), getting σi−1(k) = σj−1(k),with j − 1 ≥ 1, contradicting the fact that j was the smallest such integer.Because of the claim, it also follows that k, σ(k), σ2(k), . . . , σj−1(k) are alldistinct. Hence, O is the set described.
Theorem 6.8 (Disjoint Cycle Representation).Every permutation of n letters is a product of disjoint cycles. Further, anytwo decompositions of into such a product give the same disjoint cycles.
Proof: If σ is a permutation of N = 1, 2, . . . , n , then N is a dis-joint union of equivalence classes. Each of these classes is of the form k, σ(k), σ2(k), . . . , σj−1(k) by the lemma. Let the disjoint equivalenceclasses be k1, σ(k1), σ2(k1), . . . , σr1(k1) , k2, σ(k2), σ2(k2), . . . , σr2(k2) ,. . . , kq, σ(kq), σ2(kq), . . . , σrq(kq) .We now observe that
σ = (k1, σ(k1), σ2(k1), . . . , σr1(k1))(k2, σ(k2), σ2(k2), . . . , σr2(k2))
. . . (kq, σ(kq), σ2(kq), . . . , σrq(kq)),
a product of disjoint cycles.
Examples 6.9. Express the following as products of disjoint cycles:A. (1, 2, 3)(3, 4, 5)B. (1, 2)(2, 1, 3)(4, 3, 1)C. (1, 2, 3)−1(3, 5, 6)(2, 1)D. (a, b)(b, c)(c, d)(e, f)(f, g)
Definition 6.10. A cycle of length 2 is called a transposition.
Theorem 6.11 (Everything is 2-Cycles).Every permutation of 1, 2, . . . , n is a product of (not necessarily disjoint)transpositions.
Note: We can ignore cycles of length 1 in such a decomposition, or we canwrite, for example, (1) = (1, 2)(1, 2).
Lemma 6.12 (Effect of a 2-Cycle on the Number of Orbits).If τ is a transposition and σ is any permutation in Sn, then the number ofdisjoint orbits in τσ and σ differ by 1. (Note that disjoint orbits includethose of length 1.)
Proof: Let τ = (a, b). We look at two cases:
Case 1: a is in one of the disjoint cycles of σ and b is in another.Then we see in class how to glue τ to the three cycles involved to get a single
6. The Permutation Groups 25
disjoint one. In other words, multiplication by τ has decreased the numberof disjoint cycles by 1.
Case 2. a and b are in the same cycle of σ.Then we observe that
(a, b)(a, x, y, z, . . . , q, b, r, s, t, , , , k) = (a, x, y, z, . . . , q)(b, r, s, t, . . . , k).
In other words, it breaks up that cycle into two disjoint ones, thereby in-creasing the number of disjoint cycles by 1.
Theorem 6.13 (Parity is Well-Defined).Any two decompositions of σ as a product of transpositions have the sameparity. (That is, the number of transpositions is either odd for both or evenfor both.)
Proof: Suppose that the permutation σ ∈ Sn can be expressed in two waysas product of 2-cycles:σ = τ1τ2 . . . τs = ε1ε2 . . . εt.
We must show that s ≡ t mod 2. Butσ = τ1τ2 . . . τsι
where ι is the identity; ι = (1)(2)...(n). By the lemma, the number ofdisjoint cycles in σ must therefore be congruent to n+ s mod 2. Similarly,applying this fact to the second decomposition givesn+ s ≡ n+ t mod 2,
whence s ≡ t mod 2, as required.
Definition 6.14. A permutation that can be expressed as an even numberof transpositions is called an even permutation. Otherwise, it is called anodd permutation. We write
1 if σ is even,−1 if σ is odd.
Definition and Proposition 6.15.(a) The set of even permutations in Sn is a subgroup, called An, the
alternating group on n letters.
(b) |An| =|Sn|2
Exercise Set 6.1. Prove Lemma 6.5.2. Write down cycle decompositions of all 24 of the rigid motions of
the cube by regarding each rigid motion as a permutation of the 6faces. The set of these permutations is a subgroup of S6 which weshall call SCUBE. Prove or disprove: SCUBE is a subgroup of A6.
3. Represent the complete group of rigid motions of the regular tetra-hedron as a subgroup of S4. (Consider rotations about all possibleaxes. . . ) Deduce that this group is just A4.
4. (a) Prove that sgn: Sn → C2 is a group homomorphism.
6. The Permutation Groups 26
(b) Show that parity is not affected by conjugation; that is, if σ, ρ ∈Sn, then sgn(σ−1ρσ) = sgn(ρ).
5. (a) Let x = (1, 2)(3, 4) ∈ S8. Find an element a ∈ S8 such thata−1xa = (5, 6)(1, 3).
(b) Show that there is no element a ∈ S8 with a−1(1, 2, 3)a =(1, 3)(5, 7, 8). [Hint: see Exercise 4(b).]
6. (Herstein, p. 81 #11) Prove that the smallest subgroup of Sn con-taining (1, 2) and (1, 2, . . . , n) is Sn. (In other words, these generateSn.)
7. Cosets and Lagrange’s Theorem 27
7. Cosets and Lagrange’s Theorem
We now define a most peculiar relation on the elements of a group, arisingfrom a given subgroup:
Definition 7.1. If G is a group and H ⊂ G is a subgroup, define a relationon G by
a ≈H b⇔ a−1b ∈ HWe say that a is congruent to b mod H.
Now, saying that a−1b ∈ H is the same as saying that a−1b = h for someh ∈ H. In other words, b = ah for some h ∈ H. Thus:
a ≈H b⇔ a−1b ∈ H⇔ b = ah for some h ∈ H
Lemma 7.2. Congruence mod H is an equivalence relation.
What do the equivalence classes look like?
Definition 7.3. If H ⊂ G and g ∈ G, then define the left coset of Hcontaining g as
gH = gh | h ∈ H
Proposition 7.4 (Congruence mod H and Cosets).(a) The following are equivalent:
(i) a ≈H b(ii) a−1b ∈ H(iii) b = ah for some h ∈ H(iv) b ∈ aH(v) bH ⊂ aH(vi) bH = aH
(b) The left cosets aH are the equivalence classes of equivalence mod H.Thus, two left cosets are either equal or disjoint (this being true ofequivalence classes in general).
Proof: We prove (a) in class. For part (b), denote the equivalence class ofa ∈ G by [a]. One has
b ∈ [a] ⇐⇒ a ≈H b Definition of equivalence classes
⇐⇒ b ∈ aH By part (a), (i) ⇒ (iv)
whence [a] = aH. That is, the equivalence classes are just the left cosets, asrequired.
Examples 7.5.A. Cosets of 2Z in Z
7. Cosets and Lagrange’s Theorem 28
B. Cosets of 3Z in ZC. Cosets of nZ in ZD. G = C6; H = C3
E. G = Cmn; H = Cm
F. G = Dn; H = Cn
G. G = Z/6Z; H = 0, 3H. G = S1; H = C3
Lemma 7.6. If G is a finite group and H < G, then any two left cosets ofH have the same cardinality.2
Theorem 7.7 (Lagrange).Let G be any finite group and H < G. Then |H|
Note The “converse” is not true; if G is a group, and m is a divisor of |G|,then there need not exist a subgroup of order m. (See the exericse set.)
Corollary 7.8.(a) Every group of prime order is cyclic.(b) The order of every element of a finite group divides the order of the
group.(c) (a consequence of (b)) If G is a finite group, and g ∈ G has the
property that gk = e, then the order3 of g divides k.(d) If G is a finite group, and g ∈ G, then g|G| = e.
Definition 7.9. If H < G, then define G/H to be the set of left cosets ofH in G. If G is finite, then the index of H in G is defined as the numberof left cosets of H in G, and written as [G : H]. That is,
[G : H] = |G/H|
Consequences: (1) |G/H| = |G||H|
; (2) |G| = [G : H] · |H|.
Exercise Set 7.1. Prove that A4 has no subgroup of order 6, thus contradicting the
“converse” of Lagrange’s theorem.2. (a) List all the left cosets of C333 = 〈ω2〉 ⊂ C666.
(b) List all the left cosets of H = 〈b〉 ⊂ D8.(c) List all the left cosets of H = 〈a〉 ⊂ D8.
3. Let H < G. Define a relation ≈H on the elements of G by a ≈H biff ab−1 ∈ H.(a) Show that this is an equivalence relation.(b) Show that the equivalence classes are the right cosets, Hg =
hg | h ∈ H .2Two sets are defined to have the same cardinality if there is a bijection from one to
the other. (Notice that “having the same cardinality” is an equivalence relation on theclass of all sets.)
3See Exercise Set 4 # 8.
7. Cosets and Lagrange’s Theorem 29
(c) Show that, if G is finite and H < G, then every right coset ofH in G has the same number of elements every left coset of Hin G.
(d) Show that the number of right cosets of H in G is equal to thenumber of left cosets of H in G.
(e) Show that if G is abelian, then every left coset is a right coset.(f) Give an example to show that, in general, a left coset need not
equal any right coset. [Hint: Look at Exercise 2(b) above.]4. Describe the set of left cosets of S1×1 in T = S1×S1, where the
group structure on T is given by (s, r) · (s′, r′) = (ss′, rr′).5. Recall that two integers m and n are relatively prime if there exist
integers r and s such that rn + sm = 1. If m and n are relativelyprime, we write (m,n) = 1. Show that Cn = 〈ω〉 is generated by ωm
iff (m,n) = 1.6. (a) If H is a subgroup of G, its normalizer is
N(H) = a ∈ G | aHa−1 = H
(See Footnote.4) Prove that N(H) is a subgroup of G and con-tains H.
(b) If H is a subgroup of G, its centralizer is
C(H) = a ∈ G | ah = ha for every h ∈ H .
Prove that C(H) is a subgroup of G.(c) Is one of N(H) and C(H) contained in the other?
7. If H < G, let N =⋂
x∈G xHx−1. Prove that N is a subgroup of G
with the property that aNa−1 = N for every a ∈ G.8. (One of Herstein’s starred problems: p. 48 #24, but with hints
supplied to assist you. It is suggested that you challenge yourselfand avoid looking at the hints.)Suppose G is a finite group with |G| not divisible by 3.(a) Show that for every g ∈ G, there exists y ∈ G such that g = y3.
(That is, each element has a cube root!)(b) Suppose now that G also has the property that (ab)3 = a3b3
for every a, b ∈ G. Show that, for all a and g ∈ G, one hasga2 = a2g. [Hint: Choose y as in part (a), and consider theexpression (aya−1)3.]
(c) Deduce that, under the assumption given in part (b), G isabelian. [Hint: Use the property (ab)3 = a3b3 to conclude that(ba)2 = a2b2, and then apply part (b).]
9. Suppose G is a finite group with |G| not divisible by the prime p.Prove that there exists an integer M such that g(pM ) = g for everyg ∈ G, as follows:
4 aHa−1 is defined to be aha−1 | h ∈ H
7. Cosets and Lagrange’s Theorem 30
(a) Show that, for every g ∈ G, there exists a positive integer msuch that g(pm) = g. [Hint: Consider the elements g, gp, g(p2), . . .]
(b) Now prove that, if g(pm) = g, then g(pkm) = g for every positiveinteger k. Then use the fact that G is finite to obain the result.
8. Normal Subgroups and Quotient Groups 31
8. Normal Subgroups and Quotient Groups
Definition 8.1. A subgroup H < G is a normal subgroup of G if ghg−1 ∈H for every g ∈ G. We write H / G.
Examples 8.2.A. All subgroups of abelian groups are normal.B. An is a normal subgroup of Sn.C. 〈a〉 is a normal subgroup of Dn, whereas 〈b〉 is not.
Lemma 8.3 (Equivalent Definition of a Normal Subgroup).The subgroup H of G is normal iff gHg−1 = H for every g ∈ G.
Proof:⇒ If H is normal, then, by definition,
gHg−1 ⊂ Hfor every g ∈ G. Multiplying both sides by g and its inverse givesH ⊂ g−1Hg
for every g. Since this holds for every g ∈ G, we can replace g by itsinverse to obtainH ⊂ gHg−1
for every g. Putting these two inclusions together gives the result.⇐ If gHg−1 = H, then every element of the left-hand side is an element
of the right-hand side, showing the result.
Note By the proof of the above lemma, H < G iff gH = Hg for every g ∈ G.In fact:
Lemma 8.4 (Another Equivalent Definition of a Normal Subgroup).The subgroup H of G is normal iff every right coset of H in G is also a leftcoset.
Proof:⇒ If H is normal, then, by the preceding lemma, gHg−1 = H, whence
gH = Hg, so that the left coset gH is equal to the right coset Hg.⇐ Now suppose every left coset of H is also right coset. To show that H
is normal, choose g ∈ G. Then gH is a right coset. Since g ∈ gH, itmust also be an element of that right coset. But only one right cosetcan contain g, namely Hg. Hence, gH = Hg, whence gHg−1 = H,as required.
Definition 8.5. IfA andB are arbitrary subsets ofG, define their product,AB, by
AB = ab | a ∈ A, b ∈ B .Of course, this product is associative: (AB)C = A(BC), so we shall simplywrite ABC for a product of three subsets of G.
8. Normal Subgroups and Quotient Groups 32
A. If H < G, then HH = H.B. If H < G and g ∈ G, then gH = gH. We shall drop the set braces
for single elements, and write, say aHb instead of aHb.C. If H < G, and a, b ∈ G, then
(aH)(bH) = a(Hb)H
= a(bH)H (since H is normal)
= abH (since HH = H)
Thus, the product of two left cosets is another left coset:
(Ha)(Hb) = Hab.
Lemma 8.7 (Multiplying the Cosets of a Normal Subgroup).The subgroup H of G is normal iff the product of two left cosets of H isalways a left coset.
Proof:⇒ We just proved this part.⇐ In the exercise set.
Recall that, if H < G, then G/H is the set of all left cosets gH of H inG.
Lemma 8.8 (The Quotient Group).If H < G, then the multiplication of left cosets turns G/H into a group,called the quotient group.
Proof: We just check the axioms!
Note It follows from the last section that the order of the quotient groupG/H is the quotient, |G|/|H|.
Examples 8.9.A. G = Z; H = nZ (This explains the notation Z/nZ!)B. G any abelian groupC. Sn/An
D. Z[x]/ker εE. G/e is just a copy of G. We write G/e ∼= GF. G/G ∼= e.G. C6/ 1, ω2, ω4 H. (Z/6Z)/ , ,  I. GL(n,R)/SL(n,R)J. Dn/〈a〉 ∼= C2
K. (G×G)/(G× e)′
Exercise Set 8.1. Find all the normal subgroups of D18. Justify your assertions.
8. Normal Subgroups and Quotient Groups 33
2. (a) If G is a group and H is a subgroup of index 2 in G, prove thatH / G.
(b) Is the corresponding result still true for subgroups of index 3?Give a proof or counterexample.
3. Show that the intersection of normal subgroups of G is a normalsubgroup of G.
4. Give an example of a non-abelian group all of whose subgroups arenormal. [Hint: A certain group we have studied has this property.]
5. Suppose H is the only subgroup of order |H| in the group G. Showthat H / G.
6. Suppose that M and N are normal subgroups with the property theM ∩N = e. Show that, for any m ∈M,n ∈ N one has mn = nm.
7. Complete the proof of Lemma 8.7: The subgroup H of G is normaliff the product of two left cosets of H is always a left coset.
8. Prove that every quotient of a cyclic group is cyclic.9. Write down the multiplication tables for the following quotient groups,
and identify each as a familiar group:(a) C6/C3
(c) Q8/±1 (d) Q8/±1,±i
10. (a) Prove that, if H < G, then H /N(H). (See Exercise Set 7 forthe definition of N(H).)
(b) Prove that the subgroup H of G is normal iff N(H) = G.(c) Prove that, if H /G′, then G′ < N(H). (In other words, N(H)
is the maximal subgroup of G in which H is normal.)
9. Homomorphisms 34
Definition 9.1. Let G and G′ be groups. A map f : G → G′ is a homo-morphism if, for every pair of elements a, b ∈ G, one has f(ab) = f(a)f(b).
Examples 9.2.A. G any group; 1G : g → G; 1G(g) = g (identity map)B. G and G′ any groups, fe : G→ G′; fe(g) = e (trivial map)C. f : Z → Z; f(n) = mnD. f : Z → Z/nZ; f(n) = [n] (canonical quotient map)E. f : Z/6Z → Z/2Z; f [n] = [3n]F. G any group; f : G → G; f(g) = aga−1 for a fixed element a ∈ G.
(conjugation by a)G. det : GL(n,R) → R
H. ρ : Cn → GL(2,R); ρ(e2πi/n) =[cos(2πi/n) − sin(2πi/n)sin(2πi/n) cos(2πi/n)
]I. sgn: Sn → −1, 1; sgn(σ) =
1 if σ is even,−1 if σ is odd.
J. ε : R[x] → R; ε(p(x)) = p(1) (evaluation at 1)K. ρ : C3 → S3; ρ(ωr) = (1, 2, 3)r
L. If H < G, then the inclusion map ι : H → G; ι(h) = h is a grouphomomorphism.
M. The canonical projections π1 : G × G′ → G; π1(a, a′) = a, andπ2 : G×G′ → G′; π2(a, a′) = a′.
Lemma 9.3 (Elementary Properties of Homomorphisms).If f : G→ G′ is any homomorphism, then:
(a) f(eG) = eG′
(b) f(a−1) = f(a)−1 for every a ∈ G.(c) If H ⊂ G is a subgroup, then f(H) ⊂ G′ is also a subgroup.(d) If K ⊂ G′ is a subgroup, then f−1(K) ⊂ G is also a subgroup.
Definition 9.4. If f : G→ G′ is a group homomorphism, define:
Kerf = g ∈ G | f(g) = e = f−1(e)
Imf = f(g) | g ∈ G = f(G)
Ker f is called the kernel of f , and Im f is called the image of f . (Seethe figure.)
Corollary 9.5. Kerf and Imf are subgroups of G and G′ respectively.
Examples 9.6. We look at the kernels and images of all the homomor-phisms given in the above examples.
Definition 9.7. Let f : G → G′ be a homomorphism. We say that f is amonomorphism (injective) if f is 1-to-1. That is, f(a) = f(b) ⇒ a = b.f is an epimorphism (surjective) if f is onto; that is, Imf = G′. f is anisomorphism (bijective) if it is both monic and epic.
9. Homomorphisms 35
Note If f : G→ G′ is an isomorphism, then, by Theorem 2.25, f is invertibleas a map of sets.
Lemma 9.8 (Criterion for a Monomorphism).f : G→ G′ is a monomorphism iff Kerf = e.
Examples 9.9 (of monomorphisms and epimorphisms).A. If H < G, then the inclusion ι : H → G is a monomorphism. (Sub-
examples in class)B. If H / G, then the natural projection ν : G → G/H is an epimor-
phism. (Sub-examples in class)
Theorem 9.10 (Inverse of a Homomorphism).If f : G → G′ is an isomorphism, then f−1 : G′ → G is also a group homo-morphism.
Proof: All we need to show is that f−1(ab) = f−1(a)f−1(b) for everya, b ∈ G. But, since f is a homomorphism, applying f to each side of thisequation yields the same result: ab. Thus, since f is monic, the two sides ofthe equation must be equal as claimed.
Definition 9.11. The groups G and G′ are isomorphic if there exists anisomorphism φ : G→ G′. We write G ∼= G′.
More Terminology: An endomorphism on a group G is a homomor-phism G→ G′; an automorphism on G is an isomorphism G→ G.
Examples 9.12.A. Zn = Z/nZ ∼= Cn
B. M(n,m) ∼= Rmn
C. S3∼= D3
D. Z4∼= Z2 × Z2
E. Is Z ∼= Q under addition?F. If G is any group, define Aut(G), the automorphism group of G
to be the set of automorphisms G → G. Then Aut(G) is a groupunder composition.
9. Homomorphisms 36
Proposition 9.13 (Kernels are Normal).If f : G→ G′ is a homomorphism, then Kerf /G.
Proposition 9.14 (Preservation of Normalcy).If f : G→ G′ is a homomorphism, then:
(a) If H < J < G, then H /J implies f(H) / f(J).(b) If K < L < G′, then K /L implies f−1(K) / f−1(L).
We now prove the very important
Theorem 9.15 (Fundamental Homomorphism Theorem).Let f : G→ G′ be any homomorphism. Then there is a natural isomorphism
φ : G/Ker f∼=−→ Im f
Examples 9.16.A. sgn: Sn → −1, 1 B. f : Z → Cn; f(m) = ωm
C. f : S1 → S1; f(z) = z2
D. f : R → S1; f(x) = eix
E. f : R2 → S1 × S1; f(θ, φ) = (eiθ, eiφ)F. f : Dn → C2; f(arbs) = bs
G. det : GL(n,R) → R∗
Definition 9.17. A group G is simple if it has no normal subgroups exceptG and e.
Examples 9.18.A. Cp for primes pB. Sn is not simple for any n ≥ 2.C. An is simple if n exceeds 5 (Exercise set).
Exercise Set 9.1. Let G be any abelian group. Show that the map f : G→ G; f(g) =
g2 is a group homomorphism.2. Which of the following are group homomorphisms. Justify your
claims(a) f : (R,+) → (R∗,×); f(x) = ex
(b) f : Cn → Dn; f(ωr) = ar
(c) f : Q8 → C∗; f(±i) = ±i, f(±1) = ±1, f(±j) = ±i, f(±k) =±i.
(d) f : GL(n; R) → GL(n; R); f(A) = P−1AP for a fixed P ∈ GL(n; R).(e) Let G be arbitrary with a ∈ G. Define a : G → G by a(g) =
aga−1.3. Prove that the composite of any two homomorphisms (monomor-
phisms, epimorphisms, isomorphisms) is a homomorphism (monomor-phism, epimorphism, isomorphism).
9. Homomorphisms 37
4. Let V be a vector space over C, and let Aut(V ) be the set of linearisomorphisms V
∼=−→ V .(a) Verify that Aut(V ) is a group under composition of functions.(b) Assume that e1, e2, . . . , en is a basis for V , and let [f ] denote
the matrix of the linear map f with respect to this basis. Showthat the map φ : Aut(V ) → GL(n; C) given by φ(f) = [f ] is anisomorphism of groups. [You may quote any result from linearalgebra you like if you remember any! If not, seek help.]
5. Calculate Aut(Z/3Z) and Aut(Z/6Z). In general, what can you sayabout Aut(Z/nZ)?
6. Cayley’s Theorem Let G be a finite group of order n, and definea homomorphism φ : G→ Sn as follows.(a) Let g ∈ G. Define g : G → G by g(a) = ga. Verify that g is
bijective for every choice of g.(b) Show that the map θ : G→ SG given by θ(g) = g is a monomor-
phism.(c) Show that for each choice of numbering of the elements of G,
we have an isomorphism τ : SG∼=−→ Sn.
(d) Now let φ be the composite τ θ. Then φ is a monomorphismG → Sn. Deduce: Cayley’s Theorem: Every finite group oforder n is isomorphic to a subgroup of Sn.
7. Use Theorem 9.15 to produce isomorphisms as shown:(a) C6/C3
(b) Cpq/Cp∼= Cq
(c) Q8/±1,±i ∼= C2
8. Prove that the following statements are equivalent for the finitegroup G:(1) G is simple.(2) There is no epimorphism from G onto any group G′ with 0 <
|G′| < |G|.(3) If f : G → G′ is any homomorphism, then f is either trivial or
injective.9. The commutator of the elements a and b of the group G is the
element aba−1b−1 ∈ G. The commutator subgroup [G,G] of Gis the subgroup generated by all the commutators of elements of G.Prove:(a) [G,G] /G(b) G/[G,G] is abelian. It is called the abelianization of Ga.(c) If f : G → A is a homomorphism with A abelian, then here
exists a unique homomorphism φ : G/[G,G] → A such that thediagram
9. Homomorphisms 38
A - Bf
commutes (that is, φ ν = f). This property is called theuniversal property of the commutator subgroup.
10. Prove that any homomorphism f : G → G′ can be expressed as acomposite ι φ ν of three homomorphisms, where ι is a monomor-phism, φ is an isomorphism and ν is an epimorphism.
11. An (additive) abelian group G is called a torsion group if, for everyg ∈ G, one has ng = 0 for some g ∈ G. (0 is the identity in G.)In other words, every element of G has finite order. (See the firstsection on groups.) Prove that Q/Z is a torsion group, and illustratethis by finding an element of order 666 in Q/Z.
12. If we have an infinite sequence of homomorphisms fi : Gi → Gi−1, (i ≥1) of groups, define the inverse limit of the Gi by
←−limGi = (g0, g1, . . . , gn, . . . ) | gi ∈ Gi and fi(gi) = gi−1 .
(a) Show that←−limGi is a group.
(b) Show that the maps πi :←−limGi → Gi given by πi((g0, g1, . . . , gn, . . . )) =
gi is a group homomorphism.
(c) Show that the group←−lim (Z/piZ), where fi : Z/piZ → Z/pi−1Z
is given by the natural projection, is torsion free (that is, everyelement has infinite order).
13. Prove that An is simple if n exceeds 5 by doing the following exercisefrom Fraleigh, p. 153.(a) Show that An contains every 3-cycle if n ≥ 3.(b) Show that An is generated by 3-cycles if n = 3. [Hint: Compute
(a, b, c)(b, c, d) and (a, b)(b, c)](c) Fix r and s in 1, 2, . . . , n . Show that An is generated by the
particular 3-cycles (r, s, i) for 1 ≤ i ≤ n. [Hint: Compute(r, s, i)2, (r, s, j)(r, s, i)2, (r, s, j)2(r, s, i), and(r, s, i)2(r, s, k)(r, s, j)2(r, s, i) ]
(d) Let N be a normal subgroup of An for n ≥ 3. Show that if Ncontains a 3-cycle, then N = An. [Hint: Show that (r, s, i) ∈ Nimplies (r, s, j) ∈ N by computing((r, s)(i, j))(r, s, i)2((r, s)(i, j))−1.]
(e) Let N be a nontrivial normal subgroup of An for n ≥ 5. Showthat one of the following cases must hold, and conclude in eachcase that N = An:
Case 1: N contains a 3-cycle.Case 2: N contains a product of disjoint cycles, at least one of
which has length greater than 3. [Hint: If N contains the
9. Homomorphisms 39
disjoint product σ = µ(a1, a2, . . . , ar), show thatσ−1(a1, a2, a3)σ(a1, a2, a3)−1 is in N , and compute it.]
Case 3: N contains a disjoint product of the formσ = µ(a4, a5, a6)(a1, a2, a3). [Hint: Show thatσ−1(a1, a2, a4)σ(a1, a2, a4)−1 is in N , and compute it.]
Case 4: N contains a disjoint product of the form σ = (a1, a2, a3),where µ is a product of disjoint 2-cycles. [Hint: Showσ2 ∈ N , and compute it.]
Case 5: N contains a disjoint product of the formσ = µ(a3, a4)(a1, a2), where µ is a product of an evennumber of disjoint 2-cycles. [Hint: Show thatσ−1(a1, a2, a3)σ(a1, a2, a3)−1 is in N , and compute it todeduce that α = (a2, a4)(a1, a3) is in N . Using n ≥5 for the first time, choose i ∈ 1, 2, . . . , n with i 6=a1, a2, a3, a4. Let β = (a1, a3, i). Show that β−1αβα ∈ N ,and compute it.]
14. (Optional; Harder) Semidirect Products Let G and G′ be groups,and let φ : G → Aut(G′) be a homomorphism. Define a new groupG×φ G
′ as follows. As a set, G×φ G′ is just the Cartesian product
G×G′. Multiplication is given by the following strange formula:
(a, a′)(b, b′) = (ab, (φ(b−1)(a′))(b′)).
(a) Verify thatG×φG′ is a group with identity (eG, eG′) and (a, a′)−1 =
(a−1, φ(a)(a′−1)).(b) Show that, if φ(g) = 1: G′ → G′, then G ×φ G
′ coincides withG×G′.
(c) Let G = Cp; p prime, and let G′ = Cp ×Cp, the usual product.Define φ : Cp → Cp × Cp by φ(ωr)(ωs, ωt) = (ωs, ωrs+t). Showthat φ is indeed a homomorphism, and that G ×φ G
′ is a non-abelian group of order p3. [Write a = (ω, 1, 1), b = (1, ω, 1), c =(1, 1, ω). Show that ap = bp = cp = e, bc = cb, ac = ca, andaba−1 = bc.]
10. Some Structure Theorems 40
10. Some Structure Theorems
This section summarizes some results about homomorphisms and whatthey preserve.
Lemma 10.1. If φ : G∼=−→ G′ is an isomorphism, then φ induces a one-to-
one correspondence between subgroups of G and subgroups of G′. Moreover,this correspondence preserves normalcy.
Lemma 10.2. Let K be normal in G, and let ν : G → G/K be the nat-ural epimorphism. Then ν induces a one-to-one correspondence betweensubgroups of G containing K and subgroups of G/K. Moreover, this corre-spondence preserves normalcy.
Outline of Proof: Let S be the collection of subgroups of G containingK, and let T be the collection of subgroups of G/K. Define ψ : S → T byψ(H) = ν(H), and θ : T → S by θ(L) = ν−1(L). Then we check that ψ andθ are inverses. In checking this, note that ν−1ν(H) ⊃ H in general, but ifH ⊃ K, then
g ∈ ν−1ν(H) ⇒ν(g) ∈ ν(H)
⇒ν(g) = ν(h) for some h ∈ H⇒ν(h−1g) = e for some h ∈ H⇒h−1g ∈ Ker ν = K for some h ∈ H⇒g ∈ hK for some h ∈ H.
But, since H ⊃ K and H is a subgroups, hK ⊂ H, and so g ∈ H, showingν−1ν(H) ⊂ H as well. We already know that ψ and θ preserve normalcy.
Theorem 10.3 (“Second Homomorphism Theorem”).Let f : G→ G′ be any homomorphism with image I and kernel K. Then finduces a one-to-one correspondence between subgroups of G containing Kand subgroups of I. Moreover, this correspondence preserves normalcy.
Proof: We can write f as a composite
φ−−→∼= Iι−→ G′
By Lemmas 10.1 and 10.2, both ν and φ induce normalcy-preserving one-to-one correspondences of the appropriate type, so the result follows.
Lemma 10.4. If φ : G∼=−→ G′ is an isomorphism, then the one-to-one cor-
respondence in Lemma 10.1 induces isomorphisms between respective quo-tients. That is, if H /K < G, then φ induces an isomorphism K/H ∼=φ(K)/φ(H).
Lemma 10.5. Let K be normal in G, and let ν : G→ G/K be the naturalepimorphism. Then the one-to-one correspondence in Lemma 10.2 inducesisomorphisms between respective quotients. That is, if K < H /J < G,
10. Some Structure Theorems 41
then φ induces an isomorphism J/H ∼= ν(J)/ν(H) = (J/K)/(H/K). Inparticular, if H /G, then G/H ∼= (G/K)/(H/K).
Theorem 10.6 (“Third Homomorphism Theorem”).Let f : G→ G′ be any homomorphism with image I and kernel K. Then theone-to-one correspondence of the Second Homomorphism Theorem inducesisomorphisms between respective quotients. That is, if K < H /J < G, thenφ induces an isomorphism J/H ∼= f(J)/f(H) = (J/K)/(H/K).
11. Group Actions 42
11. Group Actions
From now on, we shall always write groups multiplicatively.
Definition 11.1. Let G be a group and X be a set. Then a left action ofG on X is a map
α : G×X → X
with the following properties for all g, h ∈ G and x ∈ X (where we writeα(g, x) as simply g.x or gx ):
(1) (gh).x = g.(h.x)(2) e.x = x
If X has a given (left) action by a group G, we refer to X as a (left) G-set.
Examples 11.2.A. Trivial action on any set XB. G acting on G by multiplicationC. G acting on G/H for H < GD. C3 acting on C by rotationE. Cn acting on C by rotationF. Cn acting on C by the clockwise rotationG. C2 acting in R by flippingH. Cn acting on e2kπi/n by multiplicationI. Dn acting on C by rotation and reflectionJ. Sn acting on 1, 2, . . . , n in the natural wayK. O(n,R) acting on Sn by multiplicationL. Disjoint unions of copies of G/H’s
Definition 11.3. If X and Y are G-sets, then define the product G-setX × Y as the Cartesian product of X and Y with the G-action specified byg.(x, y) = (gx, gy).
Definition 11.4. If X is a G-set and x ∈ X, then define
Gx = g ∈ G | g.x = x ,the stabilizer of x.
Lemma 11.5 (Stabilizers are Subgroups).Let X be a G-set and x ∈ X. Then Gx is a subgroup of G.
Note: The first example below will show that Gx need not be normal, butin fact can be an arbitrary subgroup of G.
Examples 11.6.A. X = G/H, x = eH
B. G = C3, X = e2nπi/3 , x = e2πi/3
C. X = G, x = any g ∈ GD. G = Sn, X = 1, 2, . . . , n , x = 1E. The trivial G-action on a set X.
11. Group Actions 43
F. C3 acting on C by rotation, x = 0, x = 1.
Definition 11.7. If Gx = e for every x ∈ X, then we say that X is a freeG-set, or that G acts freely on X.
Examples 11.8. Look at our list of G-actions and decide which are freeones.
Definition 11.9. Let G act on X. Then the orbit of x ∈ X is the subsetof X given by
G.x = g.x | x ∈ X The G-set X is called transitive if it is an orbit. In other words, if X = G.xfor some x ∈ X.
Examples 11.10. Look at the orbits of some elements in our examples ofgroup actions.
Proposition 11.11 (Orbits are Equivalence Classes).Let X be a G-set. The relation x ≈ y if y = g.x for some g ∈ G is anequivalence relation. The corresponding equivalence classes are the orbits inX; that is, [x] = G.x for every x ∈ X.
Corollary 11.12 (Decomposition of G-sets).It follows that every G-set is a disjoint union of orbits.
We’ll see what the orbits look like a little later on.
Examples 11.13 (of Orbits).A. G/HB. Look at the orbits of all the elements in the previous collection of
examples.C. G = C3, X = e2nπi/6, 0 . Write this set as a disjoint union of
Definition 11.14. Let X and Y be G-sets. Then a map f : X → Y iscalled a G-map, a G-equivariant map, or just an equivariant map if
f(g.x) = g.f(x)
for every x ∈ X and g ∈ G. (Note the analogy with scalar multiplication ofvectors and linear maps.)
Examples 11.15.A. The identity map on any G-setB. X any G-set and : X → ∗C. G = e. Then any map X → Y is automatically a G-map.D. G = C3, X = e2nπi/6 , Y = e2nπi/3 . What G-maps are possible?E. H < G, ν : G→ G/H the natural projectionF. K < H < G, ν : G/K → G/H the natural projection
11. Group Actions 44
Lemma 11.16 (Determining a G-map on an Orbit). If X is a transitiveG-set, then any G-map f : X → Y is completely determined by its value onany single point x ∈ X.
Proposition 11.17 (What Orbits Look Like).Let X be any G-set, and let x ∈ X. Then there is an invertible G-map
f : G/Gx → G.x
given by f(gGx) = g.x.
Definition 11.18. A G-equivalence is a G-map f : X → Y which happensto be invertible.
Proposition 11.19 (Inverses of G-maps).The inverse of a G-equivalence is a G-equivalence. (Proved in Exercises)
It follows that the map f in Proposition 11.17 is a G-equivalence.
Definition 11.20. Two G-sets X and Y are G-equivalent if there exists aG-equivalence f : X → Y . We writeX ∼=G Y . (Note that it is an equivalencerelation on the class of all G-sets—see the Exercises)
Note It follows from Proposition 11.17 that the orbit of any x ∈ X is G-equivalent to G/Gx. It now follows form Corollary 11.12 that every G-setis G-equivalent to a disjoint union of G-spaces of the form G/H for varioussubgroups H < G. Thus we now know what all G-sets “look like.” In otherwords, we have classified all G-sets.
We now look at how the various G/H are related to each other; forinstance, you might ask: When is G/H ∼=G G/K?
Definition 11.21. Let H < G and let X be a G-set. Define the H-fixedset, XH ⊂ X as:
XH = x ∈ X | h.x = x for each h ∈ H
Examples 11.22.A. eH ∈ (G/H)H
B. (G/H)H = NH/HC. If H < K, then XH ⊃ XK .D. x ∈ XGx for every x ∈ X.
(This is as far as we have to go for the Sylow theorems.)
Notation If X and Y are G-sets, denote the set of all G-maps X → Yby GM(X,Y ). (In other words, f ∈ GM(X,Y ) iff f is a G-map X →Y . Similarly, if H < G, then the set of all H-maps X → Y is denotedby HM(X,Y ). (This makes sense if X and Y are H-spaces.) Finally, ifH = e , we write e M(X,Y ) as M(X,Y ).
Notes 11.23.(a) M(X,Y ) is just the set of all maps f : X → Y .
11. Group Actions 45
(b) f ∈ HM(X,Y ) iff f(h.x) = h.f(x) for each x ∈ X and h ∈ H.
Proposition 11.24 (Fixed Points are Just G-maps).Let Y be a G-space. One has the following bijection of sets:
GM(G/H, Y ) ∼=G M(∗, Y H) ∼= Y H ,
where ∗ denotes a single-point set. In other words, G-maps G/H → Y arein 1-1 correspondence with the points in Y H .
Note: This implies that, to specify a G-map G/H → Y , all we need do ispick a point in Y H , and vice-versa.
Corollary 11.25. If Y H = ∅, then there are no G-maps G/H → Y .
Corollary 11.26. There exists a G-map G/K → G/H iff K is conjugateto a subgroup of H. Further, any G-map f : G/K → G/H has the formf(gK) = gaH, for every g ∈ G, where a ∈ G is such that a−1Ka ⊂ H. Wesay that a normalizes K in H. Conversely, every such a ∈ G determinesa G-map of the above form.
Corollary 11.27. G-maps G/H → G/H are in 1-1 correspondence withelements of NH/H.
Corollary 11.28 (Classification of Orbits up to G-Equivalence).There is a G-equivalence G/H → G/K iff K is conjugate to H in G.
Note: We now have the following sharper classification of finite G-sets:Any finite G-set is equivalent to a disjoint union of G/H’s, where any twosuch G/H’s are equivalent to each other iff the corresponding subgroups areconjugates of each other.
Lemma 11.29 (Burnside).Let G act on the finite set X. For each g ∈ G, let |Xg|denote the number ofelements fixed by g. Then the number of orbits in X is given by:
Proof: Let D = (g, x) ∈ G × X | g.x = x . Look at the projectionπ2 : D → X. Then the preimage of any point x is its stabilizer Gx. Further,the preimage of every point in the orbit of x ∈ X is a conjugate of Gx, andthus has the same number of elements. The preimage of the orbit of x ∈ Xtherefore has
Number in preimage × number in orbit = |Gx||G/Gx| = |G| elements!Thus, the total number of elements in D is
Number of orbits in G × number of elements per orbit = N |G|,so
|D| . . . (1)
11. Group Actions 46
Now look at the other projection, π1 : D → G. The preimage of g ∈ G underthis projection is just Xg. Hence,
|Xg| . . . (2)
Putting (1) and (2) together gives the result. Note In the above proof, Xg = X〈g〉. (Why?)
Example 11.30. Let us compute the number of 2-color paintings (red,green) of the faces of a regular tetrahedron. Here, X is the set of all of thosepaintings:
)= 6 altogether. For a given rotation g of the tetrahedron, Xg
consists of those paintings that are left fixed by g.
Exercise Set 10.1. Find Gx in each of the following cases:
(a) X = C, G = C2 acting by reflection in the y-axis, x = 0, x =i, x = 1.
(b) X = G/H, x = gH for some g ∈ G.(c) X = gHg−1 | g ∈ G , G acting by conjugation: g.(aHa−1) =
gaH(ga)−1;x = H.2. Show that, if X is a G-space, x ∈ X, and H < Gx, then x ∈ XH .3. Prove that the relation X ∼=G Y (of G-equivalence) is an equivalence
relation on the class of all G-sets.4. Prove Proposition 11.19.5. Prove or disprove each of the following claims:
(a) (X × Y )H = XH × Y H for X and Y be two G-sets(b) (X ∪Y )H = XH ∪Y H for X and Y sub-G-sets of some G-set Z(c) (X ∩Y )H = XH ∩Y H for X and Y sub-G-sets of some G-set Z(d) (G/K)H = GH/K where G acts by multiplication on the left
6. Show that G×G ∼=G G∐G
∐· · ·
∐G (|G| times;
union.)7. Show that G × (G/H) is G-equivalent to a disjoint union of |G/H|
copies of G.8. Represent the Sn-set 1, 2, . . . , n as a disjoint union of Sn-orbits of
the form Sn/H for certain H < Sn.9. A Consequence for p-Groups. Let p be a prime number. A p-
group is a finite group of order pn for some n. Let G be any p-group,and let G act on itself by conjugation: g.a = gag−1.(a) Show that the orbit of any element has cardinality a power of
p (possibly p0 = 1. Use Lagrange’s Theorem)(b) By thinking of G as a disjoint union of orbits and counting,
deduce that center of G is nontrivial.10. If α is an action ofG onX, then we get an associated map ρ(g) : X →
X for each g ∈ G given by
ρ(g)(x) = g.x
11. Group Actions 47
Prove: If α is an action of G on X, then:(a) The associated map ρ(g) : X → X is invertible for each g ∈ G,
with ρ(g−1) = ρ(g)−1.(b) Denote by Bij(X) the group, under composition, of invertible
maps on X. The function ρ : G → Bij(X) defined above is agroup homomorphism.
(c) The assignment α 7→ ρ is a 1-1 correspondence between ac-tions of G in X and homomorphisms G → Bij(X). (In otherwords, an action of G on X “is” precisely a homomorphismG→ Bij(X). )
(d) Let x ∈ X, and let Bijx(X) = f ∈ Bij(X) | f(x) = x . Showthat Bijx(X) is a subgroup of Bij(X). To what subgroup of Gdoes Bij(X) correspond?
11. Let X be an H-set for some H < G, and define a correspondingG-set as follows: Let G ×H X be the set of equivalence classes ofpairs (g, x) with g ∈ G and x ∈ X, where
(a, x) ≈ (b, y) iff a = bh and y = h.x for some h ∈ H.(In other words, we are defining (bh, x) ≈ (b, hx) for every h ∈ H.)We call G×HX the G-set induced by the H-action on X. Provethe following:(a) The relation ≈ is an equivalence relation.(b) G×H X is a G-set via the action g.[a, x] = [ga, x]. (Show that
this action is well defined.)(c) There is a natural G-surjection p : G×H X → G/H.(d) G×H ∗ ∼=G G/H and G×H H ∼=G G(e) If X is a G-set, then G×H X ∼=G (G/H)×X, with the product
G-action.(f) If f : Y → G/H is any G-surjection, then Y ∼=G G ×H X for
some H-set X. (Note: This result gives us a classification ofall surjections onto G-orbits.)
12. (Difficult) A decomposition of (G/H)× (G/K) into G-orbitsIf H and K are subgroups of G, define
HgK = hgk | h ∈ H and k ∈ K This is called the double coset associated with H and K.(a) Prove that two double cosets are either equal or disjoint.(b) Prove that (G/H) × (G/K) is a disjoint union of G-orbits of
the form G/(Hg ∩K), where Hg denotes gHg−1.(c) Deduce that (G/H)× (G/K) ∼=G
⋃HgK G/(Hg ∩K).
13. Compute the number of 6-color paintings of the faces of a cube (thatis, the number of possible configurations of the numbers on a die.)
14. Compute the number of 3-color paintings of the faces of a cube.
12. The Sylow Theorems 48
12. The Sylow Theorems
Definition 12.1. A p-group is a group of order pn for some prime p.
Lemma 12.2 (A Counting Result).Let G be a p-group and let X be a finite G-set. Then
|X| ≡ |XG| mod p
Theorem 12.3 (Cauchy).Let G be any finite group such that |G| is divisible by the prime p. Then Ghas an element of order p (and hence a subgroup of order p).
Proof: Let X = (g1, g2, . . . , gp) | gi ∈ G, g1g2 . . . gp = e . So X is the setof p-tuples in G with product e. Then we can say several things about X:
(a) |X| = |G|p−1. Why? Because there is a 1-1 correspondence betweenelements of X and the se of all (p− 1)-tuples of elements of G; youfill in the last element of G by the constraint gp = (g1g2 . . . gp−1)−1.
(b) Let σ be the cycle (1, 2, . . . , p) ∈ Sp. Then 〈σ〉 acts on X by per-muting the indices;
σ(g1, g2, . . . , gp) = (gσ(1), gσ(2), . . . , gσ(p))
= (g2, g3, . . . , gp, g1)
(c) (g1, g2, . . . , gp) is fixed by 〈σ〉 iff g1 = g2 = · · · = gp. For example,the element (e, e, . . . , e) is fixed by 〈σ〉.
(d) If any other element is fixed by 〈σ〉, then it has the form (a, a, . . . , a)with a 6= e, and so ap = e, so a must be an element of order p, givingus what we are seeking.
Thus, it suffices to show that X〈σ〉 has more than one element in it. But,by the lemma, since 〈σ〉 is definitely a p-group, we have
|X| ≡ |X〈σ〉| mod p.
That is,|G|p−1 ≡ |X〈σ〉| mod p
by fact (a). Since |G| = k.p, this gives:
(kp)(p−1) ≡ |X〈σ〉| mod p.
But p−1 is positive (p is at least 2), so that the left-hand side is divisible bya positive power of p, making it 0 mod p. Thus the right-hand side, |X〈σ〉|,is also divisible by p. Since X〈σ〉 has at least one element in it—namely(e, e, . . . , e)— it must therefore have order a positive power of p. Done.
Corollary 12.4. G is a p-group iff every element of G has order a powerof p.
12. The Sylow Theorems 49
Recall that G acts on G/H via left multiplication: g(g′H) = (gg′)H. Wealso saw that
|G/HH | = |NH/H|Now suppose H happens to be a p-group. Look at G/H as an H-set decom-posed into H-orbits. We have
G/H = O1∐O2
∐· · ·
where the Oi are orbits of magnitude > 1. But we know how big they mustbe: |H/J | for appropriate subgroups J of H. Since H is a p-group we musthave, by Lagrange, that each J is a p-group also, and so each |Oi| must bea non-zero power of p. In other words,
|G/HH | = |NH/H| ≡ |G/H| mod p
In particular, |NH/H| can’t be 1, since 1 is not divisible by p. Thus,NH 6= H.
Theorem 12.5 (Sylow’s First Theorem).Let G be a finite group of order pnm, where m is not divisible by p. Then:
(a) G contains a subgroup of order pi for every i ≤ n.(b) Every subgroup H of order pi is a normal subgroup of a subgroup of
order pi+1 for 1 ≤ i < n.
Proof:(a) By Cauchy’s theorem, G contains a subgroup of order p. We now
do induction on i, the start of induction i = 1 being true. Thusassume there is a subgroup of order pj for every j ≤ i. We must nowproduce a subgroup of order pi+1. Let H be a subgroup of order iand look at G/H. Since |G/H| = |G|/|H|, it is still divisible by p,so by the above boxed formula, |NH/H| is also divisible by p. ByCauchy’s theorem, the quotient group NH/H also has an element oforder p, and hence a subgroup K of order p. Look at J = ν−1(K),where ν : NH → NH/H is the natural quotient. But J contains Hand is a subgroup of NH. Further, ν| : J → K is epic with kernelH, so that J/H ∼= K. Since |H| = pi and |K| = p, we must have|J | = pi+1. Done.
(b) If H has order pi with 1 ≤ i < n, then we can get K < NH/H asabove, and find that J = ν−1(K) has order pi+1 and, being containedin the normalizer of H, must normalize H.
Definition 12.6. If G has order pnm where m is not divisible by p, wecall any subgroup of order pn a Sylow-p-subgroup of G. Thus Sylow-p-subgroups are maximal p-subgroups of G, and always exist by the theorem.
Theorem 12.7 (Sylow’s Second Theorem).Let G be finite. Then any two Sylow p-subgroups of G are conjugate.
12. The Sylow Theorems 50
Proof: Let H and K be two Sylow p-subgroups of G, and let H act onG/K by left multiplication. Then, by Lemma 12.2,
|(G/K)H | ≡ |G/K| mod p.
But |G/K| is not divisible by p (since K is Sylow) and thus neither is|(G/K)H |. In particular, (G/K)H 6= ∅. This means that there is a g ∈ Gsuch that
hgK = gK
for all h ∈ H. But this means that g−1Hg ⊂ K . Since |H| = |K|, this mustbe an equality, and we are done.
Theorem 12.8 (Third Sylow Theorem).Let G be a finite group such that p divides |G|, and let np be the number ofp-Sylow subgroups of G. Then:
(a) np divides G.(b) np ≡ 1 mod p.
Proof:(a) By Theorem 12.7, all the Sylow p-subgroups are conjugate. In other
words, if G acts on the set of all Sylow p-subgroups by conjugation,then there is a single orbit. The number of things in the orbit is np.Further, if H is a Sylow p-subgroup, then GH (the stabilizer of H)is NH. Hence:
np = |G/NH|,which certainly divides |G|.
(b) Now let H be a Sylow subgroup, and let H now act on the set X ofall Sylow p-subgroups by conjugation. Then H fixes itself.Claim: H fixes nothing else. Indeed, if H did fix another beast,K, say, then H < NK, making both H and K p-subgroups of NK.But then they must be Sylow p-subgroups of NK , making themconjugate in NK. But all conjugates of K by elements of NK arethe same as K by definition! In other words, H = K as claimed.The upshot of this is that there is only one fixed point under thisaction. Thus, by Lemma 12.2 again:
np = |X| ≡ |XH | ≡ 1 mod p.
Examples 12.9.A. All Sylow subgroups of D3
B. No group of order 15 is simple, since it has only one Sylow-5 sub-group, which must therefore be normal.