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I have to solve this exercise WITHOUT Sylow theorems and Cauchy Lemma. In fact this exercise is given in the Cayley's theorem section.

let $G$ be a group of order $2^mk$, where $k$ is odd. Prove that if $G$ contains an element of order $2^m$, then the set of all elements of odd order in $G$ is a normal subgroup of $G$.

The author gives this hint

Consider $G$ as the permutations via Cayley's theorem, and show that it contains an odd permutation

now by Cayley I have an imbedding of $G$ in $S_{|G|}$ (call $\varphi$ the injective homomorphism) using the left regular representation. Now, let $g \in G$ be the element of order $2^m$. Considering that $\alpha := g \mapsto L_g \in \text{Im}\varphi$, then $\alpha$ is a regular permutation composed by $r$ disjoint $2^m$-cycles, and so $\alpha$ is odd (an odd number of odd cycles)

but then? I don't know how to use the hint that I hope I've proved:(

I checked for similar question and I didn't found any, if someone think this is a duplicate,link the question and I'll delete this immediately:)

Boris Novikov
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Riccardo
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    The composition $G\to S_{|G|}\to C_2$ is surjective (the second arrow is the parity), so $G$ is not simple (see the kernel) – user8268 Aug 29 '13 at 19:23
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    Ok, and the kernel are the even permutations? – Riccardo Aug 29 '13 at 19:31
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    This question is answered in http://math.stackexchange.com/questions/55964 (the hypothesis there is that $G$ has a cyclic Sylow 2-subgroup, but that is the same as having an element of order $2^m$). – Derek Holt Aug 29 '13 at 19:31
  • yes (hmm, it won't allow me to post a comment with 3 letters) – user8268 Aug 29 '13 at 19:33
  • @Derek Holt there is,only 1 passage in the paper linked in the question you linked ( :-) ) that is somewhat obscure without Sylow theorems. In the final part the author discuss the order of $N \cap P_2$ and he conclude directly that is $2^{n-1}$. But without Sylow, it is sufficient to use the formula $|NP_2|=\dfrac{|N| \cdot |P_2|}{|N \cap P_2|}$ to prove that the intersection has to be of order $\geq 2^{k-1}$ and so it is of this order by lagrange-like reasoning? – Riccardo Aug 29 '13 at 20:45
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    Yes, in the notation of the paper, the kernel $N$ of the map has order $2^{n-1}$, and $N \cap P_2$ has order a power of 2 and divides $|N|$, so it cannot be larger than $2^{n-1}$ by Lagrange. – Derek Holt Aug 29 '13 at 22:48

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