Exercise 1.1.21 (If $|x|$ is odd, then $x = (x^2)^k$ for some $k$)

Question

Let $G$ be a finite group and let $x$ be an element of $G$ of order $n$. Prove that if $n$ is odd, then $x = (x^2)^k$ for some $k$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Since $n$ is odd, there is some integer $k$ such that $n = 2k-1$. Then $x^{2k-1} = 1$, so by Exercise 19, $x^{2k} x^{-1} = 1$. This shows that
$$x = (x^{2})^k$$
for some integer $k$.

\end{proof}

Exercise 1.1.21 (If $|x|$ is odd, then $x = (x^2)^k$ for some $k$)

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Exercise 1.1.21 (If $|x|$ is odd, then $x = (x^2)^k$ for some $k$)

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