A semigroup is a set $S$ with an associative operation, which we will write as multiplication. That is, $x(yz) = (xy)z$ for all $x, y, z \in S$.
a) Suppose that a finite semigroup $S$ has the property that for all $x$ there is an integer $n > 1$ such that $x^n = x$. Is it true that $S$ is, in fact, a group?
b) Suppose that a finite semigroup $S$ has the property that for all $x$ there is an integer $n$ such that $x^{n+1} = x^n$. Show that the only subsets of $S$ which form a group are of size $1$.
I am interested in how people answer this question. I am having difficulty myself.
Answer to (a) is negative. Take the semigroup $\{1,a,a^2\}$ with $a^3 = a$ (and hence $a^4 = a^2$).
Answer to (b) is a bit more difficult. Let $G$ be a subset of $S$ forming a group and let $e$ be the identity of this group. Let $x \in G$ and let $n > 0$ be such that $x^{n+1} = x^n$. Since $G$ is a finite group, we also have $x^{|G|} = e$, whence $x^{n|G|} = e$. Now $x^{n|G|} = x^{n|G|+1}$, that is $e = ex = x$. Thus $G$ is trivial.
The class of finite semigroups defined by your property are called aperiodic.