Solve $x^n=1$ in a monoid.

Question

Let $(X, *)$ a monoid with identity $e$. So can the equality $$ x^n=e $$ hold for some $n\ge 1$ when $x$ is not equal to $e$? If this can be true what is an example? If this is not true how prove it?

Answer 1

Just take the set of all complex numbers on the unit circle: $U=\{z\in\mathbb C\mid|z|=1\}$, with ordinary multiplication. This is a monoid. (Right?) The role of $e$ is played by $1$. Do equations $z^n=1$ have solutions other than $z=1$?

Now take $(0,+\infty)\subset\mathbb R$ ("positive real numbers") with ordinary multiplication. This is a monoid and $1$ is again in the role of $e$. Does $x^n=1$ imply $x=1$ there?

Answer 2

Since every abelian group is equipped with a commutative multiplication and an identity, it suffices to consider a finite abelian group. Just as Brevan has mentioned, we can consider $\mathbb{Z}/2\mathbb{Z}$, or the cyclic group of order 2 which contains the identity and only one non-trivial element $1$. Since this group is cyclic we must have $1+1=0$.