Subgroup generated by image of bihomomorphism is abelian

Statement

Suppose $G, H, M$ are groups (with some of them possibly being equal). Suppose $f : G \times H \to M$ is a bihomomorphism. Then, for any $g_{1}, g_{2} \in G$ (possibly equal, possibly distinct) and for any $h_{1}, h_{2} \in H$ (possibly equal, possibly distinct), the elements $f (g_{1}, h_{1})$ and $f (g_{2}, h_{2})$ of $M$ commute with each other, i.e.:

$f (g_{1}, h_{1}) f (g_{2}, h_{2}) = f (g_{2}, h_{2}) f (g_{1}, h_{1})$

Thus, the subgroup of $M$ given by:

$⟨ f (g, h) ∣ g \in G, h \in H ⟩$

is an abelian group.

Proof

Given: Bihomomorphism $f : G \times H \to M$ of groups, $g_{1}, g_{2} \in G$ , $h_{1}, h_{2} \in H$

To prove: $f (g_{1}, h_{1}) f (g_{2}, h_{2}) = f (g_{2}, h_{2}) f (g_{1}, h_{1})$

Proof: Consider $f (g_{1} g_{2}, h_{2} h_{1})$ . This can be expanded in two ways:

One way is to first split the left argument and then split the right argument:

$f (g_{1} g_{2}, h_{2} h_{1}) = f (g_{1}, h_{2} h_{1}) f (g_{2}, h_{2} h_{1}) = f (g_{1}, h_{2}) f (g_{1}, h_{1}) f (g_{2}, h_{2}) f (g_{2}, h_{1})$

The other way is to first split the right argument and then split the left argument:

$f (g_{1} g_{2}, h_{2} h_{1}) = f (g_{1} g_{2}, h_{2}) f (g_{1} g_{2}, h_{1}) = f (g_{1}, h_{2}) f (g_{2}, h_{2}) f (g_{1}, h_{1}) f (g_{2}, h_{1})$

Equating the two results, we get:

$f (g_{1}, h_{2}) f (g_{1}, h_{1}) f (g_{2}, h_{2}) f (g_{2}, h_{1}) = f (g_{1}, h_{2}) f (g_{2}, h_{2}) f (g_{1}, h_{1}) f (g_{2}, h_{1})$

Canceling the left-most and right-most term on both sides give us:

$f (g_{1}, h_{1}) f (g_{2}, h_{2}) = f (g_{2}, h_{2}) f (g_{1}, h_{1})$

as desired.