Induced Group Product is Homomorphism iff Commutative

Theorem
Let $$\left({G, \circ}\right)$$ be a group.

Let $$H_1, H_2$$ be subgroups of $$G$$.

Let $$\phi: H_1 \times H_2 \to G$$ be defined such that $$\forall \left({h_1, h_2}\right) \in H_1 \times H_2: \phi \left({\left({h_1, h_2}\right)}\right) = h_1 \circ h_2$$.

Then $$\phi$$ is a homomorphism iff every element of $$H_1$$ commutes with every element of $$H_2$$.

Corollary
Let $$\phi: G \times G \to G$$ be defined such that $$\forall a, b \in G: \phi \left({\left({a, b}\right)}\right) = a \circ b$$.

Then $$\phi$$ is a homomorphism iff $$G$$ is abelian.

Proof
We have $$\left({h_1, h_2}\right) \circ \left({k_1, k_2}\right) = \left({h_1 \circ k_1, h_2 \circ k_2}\right)$$.


 * Suppose $$\phi$$ is a homomorphism. Then:

$$ $$ $$ $$

This follows whatever $$k_1$$ and $$h_2$$ are, and so in order for $$\phi$$ to be a homomorphism, every element of $$H_1$$ must commute with every element of $$H_2$$.


 * Now suppose that every element of $$H_1$$ commutes with every element of $$H_2$$.

Let $$\left({h_1, h_2}\right), \left({k_1, k_2}\right) \in H_1 \times H_2$$. Then:

$$ $$ $$ $$

Thus $$\phi$$ is shown to be a homomorphism.

Proof of Corollary
$$G$$ is a Subgroup of Itself, so replace $$H_1$$ and $$H_2$$ with $$G$$ in the main result.