# Prove that if $x^2=e$ then order of $x$ is $1$ or $2$

Let $(G,*)$ be a finite group and $x$ an element of $G$.
Prove that if $x^2=e$ then the order of $x$ can be only $1$ or $2$?

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Can you reformulate the question it does not seem to make sense this way. Maybe it's just my poor english... – Leo Aug 15 '13 at 12:10
Sam157: I've tried to reformulate the question the way I understand it. (And, judging by the answers, some other people understood it the same way.) Please, check whether this is what you had in mind. – Martin Sleziak Aug 15 '13 at 13:50

You can consider two points as follows:

• Definition: The order of $x$ in group $G$ is the least positive integer $n$ such that $x^n=e_G$.

• Theorem: If $\text{o}(x)=n$ and for $m$, $x^m=e_G$ then $n|m$.

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In this case we don't even need the theorem. :-) – ShreevatsaR Aug 15 '13 at 14:27
@ShreevatsaR: Yes, because the number here is 2 but I wanted him/her to be sure enough. – Babak S. Aug 15 '13 at 14:31
Nice points, @Babak! + 1 – amWhy Aug 16 '13 at 0:05

Well, the order of $x$ is defined as the least integer $n$ such that $x^n=e$. Since $x^2=e$, the order must be less than or equal to $2$, that is, $1$ or $2$.

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I'm not sure I understand the question, this is the way I'm reading it: If $x^2=e$, is the order of $x$: $1$ or $2$? If this is it, then it depends on what is $x$:

We know that a group has a single unit, and being of order 1 means $x^1=x=e$, and therefore $e$ is the only element of order $1$, if $x\neq e$, then the order must be two (order of the subgroup it generates):

$$\left<x\right>=\left\lbrace x,x^2=e \right\rbrace$$

If the question is different, reformulate it and I will edit my answer.

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