Let M be a monoid, and define Inv(M) = {a ∈ M | a is invertible}. Prove that Inv(M) is a group.

Question

I know that it should be a group, I am just having difficulty putting a proof together. I feel like I am given a lot with the definition of any element of Inv(M) being invertible, I just seem to be running in circles about how it works. Please help!

Answer 1

Recall the definition of "identity" for a magma $(M, \cdot)$: an element $e \in M$ is an identity for $(M, \cdot)$ if $m = em = me$ for all $m \in M$. Identities, when they exist, are unique. Recall that monoids always have an identity (and the operation is associative).

Now, recall the definition of "inverse". This only makes sense with an identity! We say an element $m \in M$ of a magma $(M, \cdot)$ with a (unique) identity $e$ (e.g. if $M$ is a monoid) has an inverse $n \in M$ if $e = mn = nm$. In a monoid, inverses must be unique too, and we usually denote the inverse of $m$ by $m^{-1}$, but I'm going to refrain from doing so here.

So, you say in the comments that you need help with closure and identity. Let's deal with closure first. In particular, we need to show that, give $m, n \in \operatorname{Inv}(M)$, that $mn \in \operatorname{Inv}(M)$. That is, if $m$ and $n$ have respective inverses $a$ and $b$, then we should be able to find an inverse for $mn$ too.

I claim that $ba$ is an inverse for $mn$. That is, we should be able to verify that $$(ba)(mn) = (mn)(ba) = e.$$ To see this, consider $$(ba)(mn) = b(a(mn)) = b((am)n) = b(en) = bn = e,$$ using respectively, associativity twice, the fact that $am = e$ (since $a$ and $m$ are inverses), the fact that $e$ is the identity, then the fact that $b$ and $n$ are inverses too.

A similar argument shows that $(mn)(ba) = e$ too. This proves, by definition, that $ba$ is the (necessarily unique) identity of $mn$. Hence, $mn \in \operatorname{Inv}(M)$, as needed. This proves closure.

Now, to prove identity, you need to establish that $\operatorname{Inv}(M)$ has an identity element (and we need to figure out what it is). There's one pretty obvious candidate for this identity: the identity $e$ for the whole monoid $(M, \cdot)$. Clearly, given any $m \in \operatorname{Inv}(M)$, we have $em = me = m$, since $e$ is an identity on all of $M$. The only question is, does $e$ belong to $\operatorname{Inv}(M)$? If so, then it is an identity for $\operatorname{Inv}(M)$.

To see $e \in \operatorname{Inv}(M)$, we just need to find an inverse for $e$, i.e. some $f \in M$ such that $ef = fe = e$. However, it is easy enough to see that $f = e$ will do the trick, as $ee = e$, by applying the definition of the identity with $m = e$. Thus, $e$ is invertible (and is its own inverse), hence $e \in \operatorname{Inv}(M)$, as required.

I'm slightly concerned about the inverses argument, since as I said (in bold), you really need an identity before you can prove inverses, but I'll leave it up to you. You know that every $m \in M$ has an inverse $n$, but do you know that $n \in \operatorname{Inv}(M)$? Have a think about how to prove this.

Answer 2

$1\in I$ by definition, so $I\neq \emptyset$. $a\in I$ implies there is $b\in M$ such that $ab = ba = 1$, but then $b\in I$, as well. Can you think of why a product of invertible elements is invertible?