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 $\theta$:
- $\tuple {\tuple {g_1, h_1}, k_1} = \tuple {\tuple {g_2, h_2}, k_2}$
- $\tuple {g_1, h_1} = \tuple {g_2, h_2}$
- $k_1 = k_2$
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:
\(\ds \map \theta {\tuple {g_1, \tuple {h_1, k_1} } \tuple {g_2, \tuple {h_2, k_2} } }\) | \(=\) | \(\ds \map \theta {\tuple {g_1 g_2, \tuple {h_1, k_1} \tuple {h_2, k_2} } }\) | Definition of Group Direct Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \theta {\tuple {g_1 g_2, \tuple {h_1 h_2, k_1 k_2} } }\) | Definition of Group Direct Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\tuple {g_1 g_2, h_1 h_2}, k_1 k_2}\) | Definition of $\theta$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\tuple {g_1, h_1} \tuple {g_2, h_2}, k_1 k_2}\) | Definition of Group Direct Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\tuple {g_1, h_1}, k_1} \tuple {\tuple {g_2, h_2}, k_2}\) | Definition of Group Direct Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \theta {\tuple {g_1, \tuple {h_1, k_1} } } \map \theta {\tuple {g_2, \tuple {h_2, k_2} } }\) | Definition of $\theta$ |
Hence, $\theta$ is a (group) homomorphism.
Hence, by definition, $\theta$ is a group isomorphism.
$\blacksquare$