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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

2 Answers 2

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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.2011-05-29
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    @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.2013-12-11
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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?2013-12-11
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    @DavidHolden As one doesn't need anti-isomorphism, it suffices to note this is an involution and count fixpoints2015-06-20
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If $G$ is a group of order $2n$, show that the number of elements of $G$ of order $2$ is odd.

Define an equivalence relation on $G$ by $x\sim y$ if and only if $x=y$ or $x=y^{-1}$.
Let $C_a := \{x \in G | a \sim x\}$.
$\#C_a = 1$ if and only if the inverse of $a$ is $a$.
$\#C_a = 2$ if and only if the inverse of $a$ is not $a$.
Note that the inverse of $a$ is unique.

Let $X := \{C_a | \#C_a = 1\} = \{C_{a_1}, \cdots, C_{a_l}\}$.
Let $Y := \{C_b | \#C_b = 2\} = \{C_{b_1}, \cdots, C_{b_m}\}$.

$C_e = \{e\}$.
So, $l \geq 1$.

Then, $\#G = l + 2 m$.
Since $\#G$ is even, $l$ must be even.

$\#C_a = 1$
$\Leftrightarrow$
the inverse of $a$ is $a$
$\Leftrightarrow$
$a^2 = e$,

and

$e^1 = e$.

So, the number of elements of $G$ of order $2$ is $l-1$.
And $l-1$ is odd.