Associativity of Group Direct Product

Theorem
The group direct product $G \times \paren {H \times K}$ is (group) isomorphic to $\paren {G \times H} \times K$.

Proof
Let $G, H, K$ be groups.

The mapping $\theta: G \times \paren {H \times K} \to \paren {G \times H} \times K$ defined as:


 * $\forall g \in G, h \in H, k \in K: \map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$

is shown to be a group isomorphism, as follows:

Injective
Let $\map \theta {\tuple {g_1, \tuple {h_1, k_1} } } = \map \theta {\tuple {g_2, \tuple {h_2, k_2} } }$.

By the definition of $\theta$:
 * $\tuple {\tuple {g_1, h_1}, k_1} = \tuple {\tuple {g_2, h_2}, k_2}$

By Equality of Ordered Pairs:
 * $\tuple {g_1, h_1} = \tuple {g_2, h_2}$
 * $k_1 = k_2$

and consequently:
 * $g_1 = g_2, h_1 = h_2$

Thus:
 * $\tuple {g_1, \tuple {h_1, k_1} } = \tuple {g_2, \tuple {h_2, k_2} }$

and so $\theta$ is injective.

Surjective
If $\tuple {\tuple {g, h}, k} \in \paren {G \times H} \times K$, then $g \in G, h \in H, k \in K$.

Thus:
 * $\tuple {g, \tuple {h, k} } \in G \times \paren {H \times K}$

and:
 * $\map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$

and so $\theta$ is surjective.

Hence, by definition, $\theta$ is bijective.

Group Homomorphism
Now let $\tuple {g_1, \tuple {h_1, k_1} }, \tuple {g_2, \tuple {h_2, k_2} } \in G \times \tuple {H \times K}$.

Then:

Hence, $\theta$ is a (group) homomorphism.

Hence, by definition, $\theta$ is a group isomorphism.