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A problem which I've found quite difficult it is the following:

Let $G$ be a group with order $2^n k$ with $k$ odd, and suppose there is an element of order $2^n$ (i.e. the Sylow-$2$ subgroup is cyclic). Prove that the elements of odd order in $G$ form a normal subgroup of order $k$.

One major difficulty is in $G$ not being abelian. I've seen one proof of it, which used induction on $n$ and the group action of $G$ on itself by multiplication, noting that a permutation $\lambda_g$ is odd iff $2^n$ divides the order of $g$.

However, I've not been able to find any other methods to prove it, even after learning more group theory than I had learned at the time. I thought about using the Sylow theorems, but that didn't result in anything.

Can anyone think of other ways to prove this result? Full proofs aren't necessary; any directions which are or could be fruitful are appreciated.

SvanN
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  • The order of an element is preserved by conjugation. – lulu Aug 24 '17 at 17:58
  • Worth noting: the product of two elements of odd order could have even order, so the set of elements of odd order is not necessarily a subgroup. – lulu Aug 24 '17 at 18:01
  • I don't think the problem is well posed. As I say, the set of elements of odd order need not even be a subgroup (though it will always be preserved by conjugation). – lulu Aug 24 '17 at 18:03
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    If you want an approach that uses some more group theory, the result follows directly from Burnside's transfer theorem, which states that if a $p$-Sylow subgroup is central in its normalizer then the group has a normal $p$-complement. That this is the case here follows from the normalizer/centralizer theorem. – Tobias Kildetoft Aug 24 '17 at 18:03
  • @lulu In this case, the element of odd order really do form a subgroup. – Tobias Kildetoft Aug 24 '17 at 18:03
  • @lulu that's why the condition that there exists an element of order $2^n$ is crucial; counterexamples if that is removed are easy to come by. – SvanN Aug 24 '17 at 18:04
  • @TobiasKildetoft You are correct. I neglected the condition that there is an element of order $2^n$. – lulu Aug 24 '17 at 18:04
  • @TobiasKildetoft I hadn't heard of that theorem before - I'll be sure to look into that! – SvanN Aug 24 '17 at 18:05
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    Here is the argument for a proof by induction. No wonder I had a deja vu -experience :-) – Jyrki Lahtonen Aug 24 '17 at 18:15
  • @JyrkiLahtonen Always hard to find those answers when they were given in a somewhat different context. – Tobias Kildetoft Aug 24 '17 at 18:20
  • @JyrkiLahtonen Hmm, somehow I did not notice that it was an answer of mine (rather than your own) you had linked until it gained an upvote just now :) – Tobias Kildetoft Aug 24 '17 at 18:37
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    @TobiasKildetoft Well, it's a very nice argument indeed! – lulu Aug 24 '17 at 18:41
  • @lulu If you mean the one in the linked answer then yes, I quite like it too. I had just apparently completely forgotten that I had given that answer. – Tobias Kildetoft Aug 24 '17 at 18:42
  • Now that you just edited the question anyway, I should point out that the usual term for numbers that are not even is "odd" rather than "uneven". – Tobias Kildetoft Sep 04 '17 at 18:01
  • @TobiasKildetoft How silly of me! Thanks! – SvanN Sep 04 '17 at 18:42

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Expanding on my comment, let us apply Burnside's transfer theorem to give a nice and conceptual proof of this (and some more general facts).

Theorem (Burnside): Let $G$ be a group with a $p$-Sylow subgroup $P$ satisfying $P\leq Z(N_G(P))$. Then $G$ has a normal subgroup $N$ of index $|P|$ (i.e. a normal $p$-complement).

Theorem (normalizer/centralizer): Let $H$ be a subgroup of a group $G$. Then $C_G(H)$ is a normal subgroup of $N_G(H)$ and $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\operatorname{Aut}(H)$.

I will lave the proofs as exercises. The second one is a good exercise while the first one is a terrible exercise as it is way too hard (the two proofs I know of either use a lot of stuff about transfer or uses a lot of character theory).

Anyway, using these, we can now show that if $p$ is the smallest prime dividing the order of the group (and in the case at hand, we are considering $p=2$ which will certainly be smallest), and if the $p$-Sylow subgroup is cyclic (which is precisely the assumption), then $G$ has a normal $p$-complement, meaning that the elements of order not divisible by $p$ form a normal subgroup.

To apply Burnside's result here, we need to show that $P\leq Z(N_G(P))$, but this is the same as saying that $N_G(P) = C_G(P)$, and we can show that by showing that $N_G(P)/C_G(P)$ is trivial.

Now, by the normalizer/centralizer theorem, this quotient is isomorphic to a subgroup of $\operatorname{Aut}(P)$, which we know has order $\varphi(|P|)$ since $P$ is cyclic. But this number is coprime to $\frac{|G|}{|P|}$, whereas the order of $N_G(P)/C_G(P)$ must divide $\frac{|G|}{|P|}$ since $C_G(P)$ contains $P$. This precisely implies that $N_G(P) = C_G(P)$ as we wanted, which finishes the proof.

Tobias Kildetoft
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