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I'm having problems proving this. The full question is:

"Let $G$ be a group which order is a pair number. Show that $G$ has an element of order $2$".

Can anyone give me a hint?

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If $x\neq x^{-1}$, remove both $x$ and $x^{-1}$ from the set $G$. Repeat as many times as you can. How many elements remain in the end? What can you say about all but one of them? –  Jyrki Lahtonen Mar 27 '12 at 19:16
Ohh, I think I got this. If it is as stated, then every element $g$ of $G$ is such that $|g|>2$. But $|g|=|g^{-1}|$, and since the neutral element $e$ of $G$ is the only one element with order 1, that leaves an odd number of elements left on the group. If we separate in pairs each element and its inverse, there will be only one element left - and this element "$h$" must be its self-inverse, that is, $h^2=e$, which proves the question. –  Marra Mar 27 '12 at 19:16
It follows from Cauchy's theorem but I like @Balin's answer better. –  MJD Mar 27 '12 at 19:17
@GustavoMarra, not only one element left. See my answer. –  lhf Mar 27 '12 at 19:45
Yeah, not only one element left. But we're supposing that every element of G has order greater than 2 (in what I said). That supposition leads to an absurd. –  Marra Mar 27 '12 at 22:43
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3 Answers

up vote 15 down vote accepted

If there is no element of order 2 then show that $G$ has odd number of elements. (Hint: think of elements and inverses)

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Hint $\ $ Inversion $\rm\:x\to x^{-1}\:$ is an involution $\rm\: (x^{-1})^{-1} = x,$ so the cycles (orbits) of this permutation partition $\rm\:G\:$ into orbits of length $2$ or $1$. Since $\rm\:|G|\:$ is even so too is the number of length $1$ orbits, i.e. fixed points $\rm\:a = a^{-1};\:$ these include $\rm\:a = 1,\:$ hence, having even cardinality, must include at least one other fixed-point, necessarily of order $2$ by $\rm a^{-1}=a\:$ $\Rightarrow$ $\rm\:a^2 = 1$ but $\rm\:a\ne 1$.

For an analogous application of orbit decomposition (without parity) see my prior answer today on Wilson's theorem for groups.

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Partition $G = E \cup A \cup B$, where $E=\{e\}$, $A=\{x \in G : x = x^{-1}, x\ne e\}$, $B=\{x \in G : x \ne x^{-1}\}$. Then $E$ has one element and $B$ has an even number of elements because $B$ is invariant under inversion. Since $G$ has an even number of elements, $A$ must have an odd number of elements. In particular, $A$ has at least one element. Every element in $A$ has order 2.

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