M is a commutative monoid with identity element denoted by e.
U(M) is the set of all invertible elements of M.
Let a∈M and b∈M. Prove that if ab∈U(M), then a∈U(M).
I'm not sure if the way I approached this question is legitimate or whether there's a clearer way to show it.
Method:
Let ab ∈ U(M).
Hence $(ab)^{-1}$ ∈ U(M).
⇒ $b^{-1}a^{-1}$ ∈ U(M)
⇒ (ab)($b^{-1}a^{-1}$) ∈ U(M)
⇒ $aa^{-1}$ ∈ U(M)
Similarly for ba ∈ M we have:
$(ba)^{-1}$ ∈ U(M)
⇒ $a^{-1}b^{-1}$ ∈ U(M)
⇒ $a^{-1}b^{-1}$(ba) ∈ U(M)
⇒ $a^{-1}$a ∈ U(M).
As a ∈ M has a right inverse and left inverse in U(M), we can conclude that a ∈ U(M).
Suppose that $ab$ has an inverse $c$. Then $(ab)c = 1$, whence $a(bc) = 1$. Since $M$ is commutative, we also have $(bc)a = 1$ and thus $bc$ is an inverse of $a$.