In an arbitrary group with identity element $e$, the equation $x^2 = e$ always has one solution, namely, $x = e$. Is there always a second (different) solution? Prove or give a counterexample.
The problem I have with this question is understanding what my professor is trying to say simply because he did not describe what the binary operation of this group is. $e$ is different for any binary operation, example for multiplication it is 1 and addition is 0. Either way, 1 or 0 works in this case, namely the case where $x = e$ , I can't think of any identities in any group that this would not work.
So now, the real question is:
Is there always a second (different) solution? Either i'm misunderstanding his question, but his arbitrary group only had one solution, so I could just be sneaky and use that as my counterexample?
If I narrow it down, I can say, the group of integers under multiplication where the identity is 1, for example: if $x=-1$ or $x=1$ then the only possible case is when $x=e$ or $x=1$ is when $x=1$ since $-1 \neq e$
So my question is can someone dissect what my professor is trying to say in this problem because it is rather confusing. Thanks.