Internal and External Group Direct Products are Isomorphic

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Theorem

Let $G$ be a group whose identity is $e$.


Then $G$ is the (external) group direct product of $G_1, G_2, \ldots, G_n$ if and only if $G$ is the internal group direct product of $N_1, N_2, \ldots, N_n$ such that:

$\forall i \in \N_n: N_i \cong G_i$

where:

$\cong$ denotes (group) isomorphism
$\N_n$ denotes $\set {1, 2, \ldots, n}$


Proof

Necessary Condition

Let $G$ be the external direct product of groups $G_1, G_2, \ldots, G_n$.

For all $i \in \N_n$, let $N_i$ be defined as the set:

$N_i = \set e \times \cdots \times \set e \times G_i \times \set e \times \cdots \times \set e$

of elements which have entry $e$ everywhere except possibly in the $i$th co-ordinate.

It remains to be checked that:

$(1): \quad N_i$ is isomorphic to $G_i$
$(2): \quad N_i$ is a normal subgroup of $G$
$(3): \quad$ Every element of $G$ has a unique expression:
$\tuple {g_1, \ldots, g_n} = \tuple {g_1, e, \ldots, e} \tuple {e, g_2, e \ldots e} \ldots \tuple {e, \ldots, g_n}$

as a product of elements of $N_1, \ldots, N_n$.


$N_i$ is isomorphic to $G_i$


$N_i$ is a normal subgroup of $G$


Every element of $G$ has a unique expression


Sufficient Condition

Now let $G$ be the internal group direct product of $N_1, N_2, \ldots, N_n$.

We define a mapping $\theta: G \to N_1 \times N_2 \times \cdots \times N_n$ by:

$\map \theta {g_1 g_2 \ldots g_n} = \tuple {g_1, g_2, \ldots, g_n}$


The fact that $\theta$ is a bijection follows from the definitions.


It remains to be shown that $\theta$ is a homomorphism.


The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:

$\map \theta {g_1 g_2 \ldots g_n} \map \theta {h_1 h_2 \ldots h_n} = \map \theta {g_1 g_2 \ldots g_n h_1 h_2 \ldots h_n}$


Basis for the Induction

$\map P 2$ is the case:

\(\displaystyle \map \theta {g_1 g_2} \map \theta {h_1 h_2}\) \(=\) \(\displaystyle \tuple {g_1, g_2} \tuple {h_1, h_2}\) Definition of $\theta$
\(\displaystyle \) \(=\) \(\displaystyle \tuple {g_1 h_1, g_2 h_2}\)
\(\displaystyle \) \(=\) \(\displaystyle \map \theta {g_1 h_1 g_2 h_2}\) Definition of $\theta$
\(\displaystyle \) \(=\) \(\displaystyle \map \theta {g_1 g_2 h_1 h_2}\) Internal Group Direct Product Commutativity: $h_1 g_2 = g_2 h_1$

Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\map \theta {g_1 g_2 \ldots g_k} \map \theta {h_1 h_2 \ldots h_k} = \map \theta {g_1 g_2 \ldots g_k h_1 h_2 \ldots h_k}$


from which it is to be shown that:

$\map \theta {g_1 g_2 \ldots g_{k + 1} } \map \theta {h_1 h_2 \ldots h_{k + 1} } = \map \theta {g_1 g_2 \ldots g_{k + 1} h_1 h_2 \ldots h_{k + 1} }$


Induction Step

This is the induction step:

\(\displaystyle \) \(\) \(\displaystyle \map \theta {g_1 g_2 \ldots g_{k + 1} } \map \theta {h_1 h_2 \ldots h_{k + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map \theta {\paren {g_1 g_2 \ldots g_k} g_{k + 1} } \map \theta {\paren {h_1 h_2 \ldots h_k} h_{k + 1} }\) Group Axiom $G \, 1$: Associativity
\(\displaystyle \) \(=\) \(\displaystyle \map \theta {\paren {g_1 g_2 \ldots g_k} g_{k + 1} \paren {h_1 h_2 \ldots h_k} h_{k + 1} }\) Basis for the Induction
\(\displaystyle \) \(=\) \(\displaystyle \map \theta {g_1 g_2 \ldots g_k g_{k + 1} h_1 h_2 \ldots h_k h_{k + 1} }\) Group Axiom $G \, 1$: Associativity

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{>1}: \map \theta {g_1 g_2 \ldots g_n} \map \theta {h_1 h_2 \ldots h_n} = \map \theta {g_1 g_2 \ldots g_n h_1 h_2 \ldots h_n}$

So $\theta$ is a homomorphism.

$\blacksquare$


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