Proving that in a finite set $S$, there is an element $a \in S$ such that $a*a=a$ where $*$ is an associative binary operation on $S$, and $S$ is closed under $*$. How would I start a proof for this question? I know it is true for a set of size two:
Say $S=$ {a,b}. Then if $a*a=b$ and $b*b=a$, we can substitute to get $a*(b*b)=b \iff(a*b)*b=b$.
But since $b*b=a$, then we know that $(a*b)\neq b$. Since there are only two elements, then $(a*b)= a$. Multiplying $a$ to the left side of both: $a*(a*b)=a*a \iff a*(a*b)=b \iff (a*a)*b=b \iff b*b=b $ Which is a contradiction.
How could I extend this to any finite set? Induction? How would you begin induction on this?
