Question
Let $G$ be a group and $H$ an odd-order subgroup of $G$ of index 2 . Show that $H$ contains every element of $G$ of odd order.
Step 1
Let $g \in G$ be an element of odd order, say $|g| = 2k + 1$ for some integer $k$. We want to show that $g \in H$. Show more…
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(i) Let $G$ be a group of order $2^{m} k$, 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.) (ii) Show that a finite simple group of even order must have order divisible by $4 .$
Let $G$ be an abelian group. If $H=\left\{x \in G: x=x^{-1}\right\}$, that is, $H$ consists of all the elements of $G$ which are their own inverses, prove that $H$ is a subgroup of $G$.
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