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 $x = x^{-1}$. 
 A: In $\mathbb Z_4$ take $x=2$. Then $x+x$ is $e$ but $x\ne e$, so the answer is 'no'. 
A: [$x^2=e \Rightarrow x=e$] 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 $e=x^n=x^{2m+1}=(x^2)^m \cdot x= e^m \cdot x= x$.
A: $(-1)^2=1 \phantom{                   }$.
A: 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.
A: It means, that if $x \neq e$ the $|x|=2$ (the order of $x$).
