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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?

  • 0
    Possible duplicate of [Group of even order contains an element of order 2](http://math.stackexchange.com/questions/42034/group-of-even-order-contains-an-element-of-order-2)2016-11-10

3 Answers 3

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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.