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:

$\langle f(g,h)\mid g\in G,h\in H\rangle$

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.