# Internal and External Group Direct Products are Isomorphic

## 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 $N_1, \ldots, N_n$.

### 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 $\text G 1$: 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 $\text G 1$: 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.

$\blacksquare$