Let E be a non empty finite set, and let ∗ be an associative law on E. Show that there exists an element s ∈ E such that s ∗ s = s.

Question

right now my tactics are to use some kind of proof by induction/contradiction, but have not really got anywhere. Any suggestion is appreciated

Answer 1

In standard terminology, what you are asking for is a proof that every finite semigroup (= non-empty set with an associative binary operation) contains an idempotent (= element $e$ such that $e^2=e$). [Here and below, I omit the "$*$"-symbol for the binary operation.]

Let $a$ be any element in $E$. Since $E$ is finite, the list of elements $a,a^2, a^3,\ldots$ must contain repetitions. Let $m$ be the smallest positive integer such that $a^m = a^{m+x}$ for some $x\ge1$, and let $r$ be the smallest among the numbers $x\ge1$ satisfying this equation.

The list $m,m+1,\ldots,m+r-1$ consist of precisely $r$ consequtive integer, and thus it contains and element $y$ such that $y\equiv 0$ modulo $r$. Now, using elementary arithmetics, you can show that $e := a^y$ is an idempotent, that is, $e^2 = e$.

More generally, $\{a^m,a^{m+1},\ldots, a^{m+r-1}\}\subset E$ is a subsemigroup (that is, it is closed under the binary operation), which is isomorphic to the cyclic group $(\mathbb{Z}/r\mathbb{Z},+)$. Do the arithmetics yourself, or look it up, for example in Howie's Fundamentals of semigroup theory (Oxford University Press, 1995), Section 1.2.