Actually, this question already has multiple answers on this website: using Burnside's theorem and one with induction and $p$-Sylow groups. I'm asking this question here, however, because the exercise below appears in my group theory syllabus in the third chapter, with only the following topics covered: definition of groups, many examples, subgroups, direct product, homomorphisms, generators, order, index. Thus I don't understand the two answers I found on this site and I'm looking for a more elementary approach using the topics included in the first three chapters of my syllabus.
So I'm asked the following.
Let $G$ be a finite group of order $2^tk$, $\ t,k\in\mathbb{Z}$, $k$ odd and suppose that $G$ has an element of order $2^t$. Prove that the elements of $G$ of odd order form a subgroup of order $k$ and index $2^t$ in $G$.
Everything I tried so far led me nowhere and it does not contribute anything to show this here. I hope anyone can be give me a hint or (partial) proof to get me going!
Here's a sketch of a proof, using only very basic notions.
Consider an element $g$ of order $2^t$. The image of $g$ in the regular representation must consist of $2^t$-cycles, so $k$ of them. Since $k$ is odd and $2^t$ is even, this is an odd permutation. So the elements of $G$ that correspond to even permutations in the regular representation form a subgroup $H$ of index $2$. Note that all elements of odd order of $G$ are contained in $H$.
Note that $g^2\in H$ has order $2^{t-1}$, while $|H|=k2^{t-1}$. So $H$ also satisfies the hypothesis. Simply repeat the procedure and you get a chain of subgroups, each of index $2$ in the previous one, until, you get a subgroup $N$ of index $2^t$ in $G$ and thus of order $k$, which contains all the elements of odd order in $G$. (Since $k$ is odd, by Lagrange $N$ consists of exactly the elements of odd order.)
(EDIT: I've made a small edit to remove the use of characteristic subgroups.)