# If in a group G an element $x$ has the property such that $x^2 = e$, does that entail $x = e$?

I can't seem to prove that $x = e$ from $x = x^{-1}$.

-
That’s a good thing: it’s not true! – Brian M. Scott Feb 27 '13 at 10:17
It holds if your group has odd order though, for example in $\mathbb{Z}_3$. – user1729 Feb 27 '13 at 10:40
You may upvote as many helpful answers as you'd like, and you may accept one answer per question: to accept an answer, simply click on the $\checkmark$ to the left of the answer you'd like to accept. User's who help out here don't get paid; we answer questions because we love math and want to help others understand math, as needed. One way to say "thank you" is to upvote and accept helpful answers. (Plus you get 2 reputation points for every answer you accept!) – amWhy Mar 4 '13 at 20:46

That's because it can't be done. Consider the group consisting of $-1$ and $1$ under ordinary multiplication.

There are many other examples. An element $x$ such that $x^2=e$ but $x\ne e$ is called an element of order $2$. Many groups have one or more elements of order $2$.

Consider for example the group of distance-preserving mappings from the plane to itself. Let $a$ be rotation about the origin through $180^\circ$. Then $a^2$ is the identity, but $a$ is not. Let $b$ be reflection in a certain line $\ell$. Then $b^2$ is the identity, but $b$ is not.

Or else consider the group of all permutations of the set $\{a,b,c,d,e\}$. Let $\sigma$ be the permutation that interchanges $a$ and $b$, and leaves the others alone. Then $\sigma^2$ is the identity, but $\sigma$ is not.

-
Oh. Thanks. I'm stupid haha. – user8210 Feb 27 '13 at 10:12
You just have not yet accumulated a stock of basic examples of groups, to test conjectures against. – André Nicolas Feb 27 '13 at 10:25

$(-1)^2=1 \phantom{ }$.

-
I make this CW in order to stop these stupid upvotes ... actually André's answer already contains my answer, but I didn't read it carefully enough. – Martin Brandenburg Feb 28 '13 at 0:03

In $\mathbb Z_4$ take $x=2$. Then $x+x$ is $e$ but $x\ne e$, so the answer is 'no'.

-
Or in $\mathbb Z_{2n}$ take $x=n$... – nbubis Feb 27 '13 at 10:13

[$x^2=e \Rightarrow x=e$] if the order of $x$ is odd. For let $n$ be the order of $x$ and write $n=2m+1$, assume $x^2=e$. Then $e=x^n=x^{2m+1}=(x^2)^m \cdot x= e^m \cdot x= x$.

-

It means, that if $x \neq e$ the $|x|=2$ (the order of $x$).

-