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

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

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 seen to be 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 seen to be surjective.

### 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:

 $\displaystyle \map \theta {\tuple {g_1, \tuple {h_1, k_1} } \tuple {g_2, \tuple {h_2, k_2} } }$ $=$ $\displaystyle \map \theta {\tuple {g_1 g_2, \tuple {h_1, k_1} \tuple {h_2, k_2} } }$ $\displaystyle$ $=$ $\displaystyle \map \theta {\tuple {g_1 g_2, \tuple {h_1 h_2, k_1 k_2} } }$ $\displaystyle$ $=$ $\displaystyle \tuple {\tuple {g_1 g_2, h_1 h_2}, k_1 k_2}$ $\displaystyle$ $=$ $\displaystyle \tuple {\tuple {g_1, h_1} \tuple {g_2, h_2}, k_1 k_2}$ $\displaystyle$ $=$ $\displaystyle \tuple {\tuple {g_1, h_1}, k_1} \tuple {\tuple {g_2, h_2}, k_2}$ $\displaystyle$ $=$ $\displaystyle \map \theta {\tuple {g_1, \tuple {h_1, k_1} } } \map \theta {\tuple {g_2, \tuple {h_2, k_2} } }$

showing that $\theta$ is a (group) homomorphism.

The result follows.

$\blacksquare$