Product with Inverse on Homomorphic Image is Group Homomorphism

Theorem
Let $G$ be a group.

Let $H$ be an abelian group.

Let $\theta: G \to H$ be a (group) homomorphism.

Let $\phi: G \times G \to H$ be the mapping defined as:


 * $\forall \tuple {g_1, g_2} \in G \times G: \map \phi {g_1, g_2} = \map \theta {g_1} \map \theta {g_2}^{-1}$

Then $\phi$ is a homomorphism.

Proof
First note that from Group Homomorphism Preserves Inverses:


 * $\map \theta {g_2}^{-1} = \paren {\map \theta {g_2} }^{-1} = \map \theta { {g_2}^{-1} }$

and so $\map \theta {g_1} \map \theta {g_2}^{-1}$ is not ambiguous:


 * $\map \theta {g_1} \map \theta {g_2}^{-1} = \map \theta {g_1} \paren {\map \theta {g_2} }^{-1} = \map \theta {g_1} \map \theta { {g_2}^{-1} }$

From External Direct Product of Groups is Group, $G \times G$ is a group.

Let $a_1, a_2, b_1, b_2 \in G$.

We have:

Hence the result by definition of homomorphism.