Exercise 4.2.13 ( If $|G| = 2k$ where $k$ is odd then $G$ has a subgroup of index $2$)

Question

Prove that if $|G| = 2k$ where $k$ is odd then $G$ has a subgroup of index $2$. [Use Cauchy's Theorem to produce an element of order $2$ and then use the preceding two exercises.]

Richard Ganaye · Accepted Answer

\begin{proof} 
Suppose that $|G| = 2k$ where $k$ is odd .

By Cauchy's Theorem, there is some element $x \in G$ of order $2$. Then $|x| = 2$ and $|G|/|x| = k$ is odd, so, by Exercise 11, $\pi(x)$ is an odd permutation. Then Exercise 11 proves that $G$ has a subgroup of index $2$.
\end{proof}

Exercise 4.2.13 ( If $|G| = 2k$ where $k$ is odd then $G$ has a subgroup of index $2$)

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Exercise 4.2.13 ( If $|G| = 2k$ where $k$ is odd then $G$ has a subgroup of index $2$)

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