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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!

Václav Mordvinov
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    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? – Tobias Kildetoft Mar 06 '18 at 20:01
  • 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. – Václav Mordvinov Mar 06 '18 at 20:03

1 Answers1

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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.)

verret
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  • It seems like the OP has not yet seen group actions. – Tobias Kildetoft Mar 06 '18 at 21:07
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    @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. – Tobias Kildetoft Mar 06 '18 at 21:09
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    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. – verret Mar 06 '18 at 21:10
  • Why do both answers assume that a relatively advanced result (relative to what the OP himself said he's studied so far) as Cayley's Theorem is "basic"? Not only that: both answers make relatively deep use of the structure of element in symmetric groups, whic also hasn't yet been studied by the OP... I think Tobias' comment below the question still applies: how can this exercise be solved with so few stuff as the OP says he's studied? – DonAntonio Mar 06 '18 at 23:29
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    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? – verret Mar 06 '18 at 23:57
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    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"? – verret Mar 06 '18 at 23:57
  • @verret I don't assume anything: he himself wrote that explicitly in his post, as you can easily read. That's happened to most of us now and then...And about books: Rotman's book takes it in page 52, in the obvious chapter of permutation groups and etc...even after proving the simplicity of $;A_n,,n\ge5;$ . Dummit&Foote: in the chapter of actions of groups, after seeing permutations, series (Hölder and stuff), etc. Hungerford: after actions, permutations and etc. – DonAntonio Mar 07 '18 at 00:23
  • @verret About your second comment above: nobody said it would be better to write that...and, as far as I can read, nobody did. I, as Tobias said, also don't know a proof that uses only the elements the OP wrote has studied so far. Perhaps somebody does. The answers here don't show that, imo. They use stuff more advanced than what the OP stated he's studied so far. That's all. – DonAntonio Mar 07 '18 at 00:27
  • Indeed, like DonAntonio assumes, I'm also in doubt whether this result can be proven using the topics I named. And indeed group actions and Cayley's theorem appear later in my syllabus. Like verret says, an answerer cannot know exactly what is treated in my course and what is not and thus have to make assumptions on this. – Václav Mordvinov Mar 07 '18 at 11:31
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    Also, I agree with verret that writing this answer is better than do nothing. Essentially (as far as I can judge) this answer seems the same as the one by induction I linked to in the question, but this one seems like a 'translation' to more elementary terminology so I can understand this. So thanks for this, also to Tobias who wrote the answer I linked to.

    By the way I also looked quickly this morning to the other answer that is deleted now but I wasn't able to follow that one.

     – Václav Mordvinov Mar 07 '18 at 11:34
  • By the way when you say

    "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 $H$.",

    this is where Cayley's theorem and group actions are used, implicitly, right? If not, could you elaborate on this a bit further because I get it intuitively but I wouldn't know how to make this rigorous.

    Like DonAntonio says, I'm also in doubt whether an easier solution is possible. This morning I asked my teacher but she also couldn't think of an elementary solution.

     – Václav Mordvinov Mar 07 '18 at 11:37
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    Nice answer (+1). My opinion is that this is both an elementary solution, and the best possible solution. – Stephen Mar 07 '18 at 14:39
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    @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 $H$.) – verret Mar 07 '18 at 22:27
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    @VáclavMordvinov For the first part, the regular representation gives you a homomorphism from $G$ to $Sym(G)$. To make the notation easier, let me identify $G$ with its image in $Sym(G)$. (It turns out that the homomorphism is injective, so this is an embedding, but that doesn't even matter.) Let $H=G \cap Alt(G)$. Since $|Sym(G):Alt(G)|=2$, we have $|G:H|\leq 2$. (see my answer for https://math.stackexchange.com/questions/2649635/centralizer-sizes-in-groups-and-subgroups/2649673#2649673 for example) Since we have an odd permutation, $H<G$, so in fact $|G:H|=2$. – verret Mar 07 '18 at 22:32
  • It makes way more sense now and I see that we have proved a lot of stuff we mention already in the exercises of our syllabus. I understand now, although it is a quite difficult exercise at this stage. Thanks for the help! :) – Václav Mordvinov Mar 08 '18 at 06:49