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This is an exercise from Rotman's An Introduction to the Theory of Groups:

Exercise 3.30. 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 $G$. (Hint: Consider $G$ as permutations via Cayley's theorem, and show that it contains an odd permutation.)

My attempt: Let $a\in G$ with $|a|=2^m$. Following the hint, we imbed $G\hookrightarrow S_G$ and identify $G$ with its image. The element $a$ is the product of $k$ disjoint $2^m$-cycles (since $a$ is a regular permutation) and thus $a$ is an odd permutation. Then $G$ has exactly $2^{m-1}k$ odd permutations. Any element in $G$ of odd order is an even permutation. However, the converse is false. I am unable to proceed. I think I must be missing something. Any hints will be appreciated!

Yuxiao Xie
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1 Answers1

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Every element of odd order is an even permutation. So they all lie in the subgroup $H$ of $G$ consisting of those elements mapping to an even permutation in $S_G$. Then $|H|=2^{m-1}k$. Moreover $H$ has an element of order $2^{m-1}$ (why?). So inductively we may assume the elements of odd order of $H$ form a subgroup of $H$, etc.

Angina Seng
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