The set of invertible elements of a monoid is closed under multiplication

Question

Let $M$ be a monoid and let $U(M)$ be the set of invertible elements of $M$. How can I prove that $U(M)$ is closed under the binary operation on $M$, i.e., that that $a \in U(M)$ and $b \in U(M)$ implies that $ab \in U(M)$?

Answer 1

Hint: if $a^{-1}$ is the inverse of $a$ and $b^{-1}$ is the inverse of $b$, compute $(ab)(b^{-1}a^{-1})$ and $(b^{-1}a^{-1})(ab)$. What can you conclude?