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I am working on the following problem from group theory:

If $G$ is a group of order $2n$, show that the number of elements of $G$ of order $2$ is odd.

That is, for some integer $k$, there are $2k+1$ elements $a$ such that $a \in G,\;\; a*a = e$, where $e$ is the identity element of $G$.

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    Your final statement is not *quite* accurate, because there is an element $a$ with $a*a=e$ that is *not* of order $2$, to wit $e$ itself. In fact, in the situation you describe, the number of elements $a$ of $G$ that satisfy $aa=e$ will be even, because there will be an odd number of elements of order $2$, and you'll also have the identity element.2011-05-29
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    @Arturo: One might use this to generalize the result to say that the number of square roots of the identity element always has the same parity as the order of the group.2011-05-29
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    @joriki: Indeed. Of course, even more is true: Frobenius's Theorem gives that the number of solutions to $x^n=1$ in a finite group $G$ is always a multiple of $\gcd(n,|G|)$. For $n=2$, the theorem implies the case of $|G|$ even, while Lagrange implies the case of $|G|$ odd.2011-05-30

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