A simple interesting result in Group Theory on the existence of (normal) subgroup is following:

If $|G|=2m$, $m$ odd, then $G$ has a (normal) subgroup of order $m$.

This theorem, as a problem/question, has been appeared on StackExchange several times.

I know the proof, and when I saw the proofs in all the appeared answers in StackExchange, it is noticed that the answers are same.

So, I asked myself: Is there a different proof of this theorem?

I tried the following way, and feel that, the following arguments are very natural ones; but I couldn't complete.

Suppose, we know there is subgroup $H$ of order $m$ (odd). What can be said about this subgroup? Since $[G\colon H]=2$, the square of every element of $G$ will be in $H$. Conversely, if $x$ is any element of $H$, then being element of odd order, $x$ is a square.

It follows that $H$ is precisely the set of squares in $G$.

Now, I tried to prove the theorem in following way: it suffices to consider a subset $$S=\{g^2\colon g\in G\}$$ and prove that $S$ is a subgroup. In fact, it is sufficient to prove that $S$ is closed under product (since it is subset of finite group).

Question: Given $|G|=2m$, $m$ odd, can it be proved that the subset of squares in $G$ is closed under product, without using the arguments (or Cayley's theorem) of the quoted theorem above?

  • $\begingroup$ If is sufficient to prove : ord(a) odd and ord(b) odd $\implies$ ord(ab) odd because the squares of $G$ are the elements of odd order of $G$. $\endgroup$ – Peter Mar 31 '16 at 12:01

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