I have to show the following.
Let $M$ be a monoid. If $M \to M$, $f: a \mapsto a^2$ is a homomorphism, then $M^\times$ is commutative.
So, if $f$ is a homomorphism, then for all $a,b \in M^\times$ we have $a \circ b = b \circ a$, where $\circ$ is the operation of the monoid $M$.
I know that for $M$, which is a set equiped with a operation $\circ$,the following axioms are satisfied:
- Closed under operation
- Associativity
- Neutral element
Given two structures $(A, \circ)$ and $(B, \ast)$, a homomorphism $A \to B$ is a map $f: A \to B$ for which holds $f(a \circ b) = f(a) \ast f(b), \ \ a \in A$ and $b \in B$. In my case, both $A$ and $B$ are the monoid $M$ and the operation is $\circ$ (because the monoid operation is not explicitly defined). So, the homomorphism $f(a \circ b) = f(a) \circ f(b) $ looks like this:
$$(a \circ b)^2 = (a)^2 \circ (b)^2$$
$M^\times$ is the set of all invertible elements of $M$, so $M^\times = \{a \in M : \exists \ a^{´} \in M | \ a \circ a^{´} = e = a^{´} \circ a \}$.
My first question is how to approach this exactly? I don't see how commutativity follows from the homomorphism.
And secondly, do I have to show only that if $f$ is a homomorphism then $M^\times$ is commutative or do I also have to show the other direction, i.e. if $M^\times$ is commutative then $f$ is a homomorphism?
Assume $f\colon M\to M, m\mapsto m^2$ is a homomorphism. Let $a, b\in M^\times$ be arbitrary. We wish to show $ab=ba$. Applying $f$ to $ab$ gives us the equation $abab = (ab)^2=a^2b^2$. Multiply from the left by $a^{-1}$ and from the right by $b^{-1}$ to obtain $ba=ab$, as desired. Note, however, that this does not work if $a\not\in M^\times$ (or $b$), since then it will not have an inverse. An immediate corollary is that if $M$ is not just a monoid but a group (so that $M^\times = M$), then $f$ is a homomorphism iff $M$ is commutative. $(\ ^\circ\smile^\circ)$