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I am working on the following problem from group theory:

If $G$ is a group of order $2n$ (i.e., the cardinality of the set $G$ is $2n$), 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. –  Arturo Magidin May 29 '11 at 22:55
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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. –  joriki May 29 '11 at 23:25
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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. –  Arturo Magidin May 30 '11 at 2:00

1 Answer 1

Hint. Define an equivalence relation on $G$ by $x\sim y$ if and only if $x=y$ or $x=y^{-1}$. Then remember that an equivalence relation partitions a set.

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Thank you Arturo. This helped to make the problem very clear. –  Jacobsen May 29 '11 at 23:23
    
@Arturo - i remember encountering this problem a few months ago as an exercise in prof Milne's notes on group theory chapter 1 - which is as far as i got - had me stumped until i twigged the obvious. your answer is a very elegant way of expressing the matter. as yet i have not learned how to vote, but i would certainly give it a plus if i knew how. –  David Holden Dec 11 '13 at 7:57
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PS might one also express the same idea in terms of the kernel of the anti-isomorphism $g \rightarrow g^{-1}$, or is there a taboo on using this naughty mapping? –  David Holden Dec 11 '13 at 8:14

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