Question: I've struggled with this one but not too many clues, it says:

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> Let $G$ be a group which all subgroups are normal to $G$, and $a,b\in G$, then $ba=a^jb$ for some $j$.
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It doesn't say what $j$ is but according to the book it shouldn't be a element of $G$ so $j$ should be an integer. First we have a subgroup $H$ of $G$ contains $x^{-1}ax$ for all $x\in G$ and $a\in H$, besides that I tried contradiction and substitution, and I think the approach is to take $G$ itself as $H$ but I don't get the proof, any hint?

Answer: 
Since the subgroup of $G$ generated by $a$ is normal in $G$,then $bab^{-1}$ belongs to this subgroup, i.e., there is $j$ such that $bab^{-1}=a^j$ or $ba=a^jb$.
Hint. The subgroup of $G$ generated by $a$ consists of all elements of the form $a^j$.