Associativity of Group Direct Product

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

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

The mapping $$\theta: G \times \left({H \times K}\right) \to \left({G \times H}\right) \times K$$ defined as:

$$\forall g \in G, h \in H, k \in K: \theta \left({\left({g, \left({h, k}\right)}\right)}\right) = \left({\left({g, h}\right), k}\right)$$

is to be shown to be a group homomorphism, and that $$\theta$$ is bijective, as follows:

Injective
Let $$\theta \left({\left({g_1, \left({h_1, k_1}\right)}\right)}\right) = \theta \left({\left({g_2, \left({h_2, k_2}\right)}\right)}\right)$$.

By the definition of $$\theta$$, $$\left({\left({g_1, h_1}\right), k_1}\right) = \left({\left({g_2, h_2}\right), k_2}\right)$$.

By the definition of ordered pairs, $$\left({g_1, h_1}\right) = \left({g_2, h_2}\right), k_1 = k_2$$.

And so on: $$g_1 = g_2, h_1 = h_2$$

Thus $$\left({g_1, \left({h_1, k_1}\right)}\right) = \left({g_2, \left({h_2, k_2}\right)}\right)$$ and $$\theta$$ is seen to be injective.

Surjective
If $$\left({\left({g, h}\right), k}\right) \in \left({G \times H}\right) \times K$$, then $$g \in G, h \in H, k \in K$$.

Thus $$\left({g, \left({h, k}\right)}\right) \in G \times \left({H \times K}\right)$$ and $$\theta \left({\left({g, \left({h, k}\right)}\right)}\right) = \left({\left({g, h}\right), k}\right)$$ and $$\theta$$ is seen to be surjective.

Homomorphism
Now let $$\left({g_1, \left({h_1, k_1}\right)}\right), \left({g_2, \left({h_2, k_2}\right)}\right) \in G \times \left({H \times K}\right)$$. Then:

$$ $$ $$ $$ $$ $$

thus showing that $$\theta$$ is a homomorphism.

The result follows.