# External Direct Product of Abelian Groups is Abelian Group

## Theorem

Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups.

Then the group direct product $\struct {G \times H, \circ}$ is abelian if and only if both $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are abelian.

### General Result

The external direct product of a finite sequence of abelian groups is itself an abelian group.

## Proof

Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups whose identities are $e_G$ and $e_H$ respectively.

From External Direct Product of Groups is Group, $\struct {G \times H, \circ}$ is indeed a group whose identity is $\tuple {e_G, e_H}$.

Suppose $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are both abelian.

Then from External Direct Product Commutativity, $\struct {G \times H, \circ}$ is also abelian.

Now suppose that $\struct {G \times H, \circ}$ is abelian.

Then:

 $\ds \tuple {g_1 \circ_1 g_2, e_H}$ $=$ $\ds \tuple {g_1 \circ_1 g_2, e_H \circ_2 e_H}$ Definition of $e_H$ $\ds$ $=$ $\ds \tuple {g_1, e_H} \circ \tuple {g_2, e_H}$ Definition of Group Direct Product $\ds$ $=$ $\ds \tuple {g_2, e_H} \circ \tuple {g_1, e_H}$ as $\struct {G \times H, \circ}$ is abelian $\ds$ $=$ $\ds \tuple {g_2 \circ_1 g_1, e_H \circ_2 e_H}$ Definition of Group Direct Product $\ds$ $=$ $\ds \tuple {g_2 \circ_1 g_1, e_H}$ Definition of $e_H$

Thus:

$g_1 \circ_1 g_2 = g_2 \circ_1 g_1$

and $\struct {G, \circ_1}$ is seen to be abelian.

The same argument holds for $\struct {H, \circ_2}$.

$\blacksquare$