Let $G$ be a finite group of even order. Prove that there exists an element $a \in G$ of order $2$.
My attempt:
Since the order of $a$ is even, $o(a)$ = the least positive integer $n$ such that $a^{2n}$=e.
Let $o(a)= 2n$
$a^{2n} = (a^2)^n = a^n = e$
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1Nope. $o(a)$ is the least positive intger $n$ such that $a^n=e$. The order of $a$ need not be even a priori – Hagen von Eitzen Oct 02 '16 at 17:36
2 Answers
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The map $x\mapsto x^{-1}$ is an involutionary map of $G$ to itself, hence must have an even number of fixpoints. As $e$ is one fixpoint, ther must exist at least one more fixpoint $a$. Then $a=a^{-1}$ and $a\ne e$, so $o(a)=2$.
Hagen von Eitzen
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Let $a \in G$. Then $o(a) = 2k$ with $k\in \mathbb{N}$. Then $a^k \in G$ and $(a^k)^2 = e_G$. Since $a_k \neq e_G$, we conclude that $o(a_k) = 2$, therefore finishing the proof.
Hermès
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