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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!

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    $\begingroup$ Let me just remark immediately (as the answerer for the other methods) that I have no idea off the top of my head how one might do this with nothing more than the given tools. Has the exercise been marked in some way as challenging? $\endgroup$ – Tobias Kildetoft Mar 6 '18 at 20:01
  • $\begingroup$ No, but none of the exercises are marked as challenging and there are some that are really difficult or require quite difficult theorems. By the way I state the topics included in my post but this also includes many related theorems. I would also appreciate another approach then the two in the other answers, because it might be helpful for me to understand this result. $\endgroup$ – Václav Mordvinov Mar 6 '18 at 20:03
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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.)

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    $\begingroup$ @Stephen Sure, but they had already seen the answer in the link which is practically the same as this one, so clearly some part of it was not familiar. $\endgroup$ – Tobias Kildetoft Mar 6 '18 at 21:09
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    $\begingroup$ I was assuming that the mention of Sylow subgroups (and even Hall subgroups) in those answers was throwing them off. Things like Cayley's Theorem are usually seen much earlier than Sylow (which is often not even included in a first class in group theory) and, in any case, it is actually very easy to show what is needed here from scratch (as Stephen said). But I also agree with both of you, these answers are essentially all the same, I just tried to avoid more advanced terminology. $\endgroup$ – verret Mar 6 '18 at 21:10
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    $\begingroup$ Speaking of assuming, why do you assume that the OP has not studied these? All we've been given is a list of words, something like titles of sections, and it's hard to know exactly what that means. In most basic group theory classes, Cayley's Theorem is treated very early. In fact, I just had a look, and in the first two books I checked, it is handled in the chapter on homomorphisms, which the OP said they have seen. Similarly with the parity of a permutation. So who knows? Why don't we wait for the OP to tell us if they are satisfied with these answers, and clarify some parts if necessary? $\endgroup$ – verret Mar 6 '18 at 23:57
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    $\begingroup$ I did my best to write the most elementary answer I know. As Stephen, I doubt if there exists a significantly easier one. Would it have been better if nobody wrote anything except "it is impossible to write a proof using only the list of words you gave us"? $\endgroup$ – verret Mar 6 '18 at 23:57
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    $\begingroup$ @VáclavMordvinov: the second part is easier. If $H$ has index $2$, then any element not in $H$ must have even order. (Its powers will alternate between $H$ and the other coset, so only even powers will be $H$.) $\endgroup$ – verret Mar 7 '18 at 22:27

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