External Direct Product of Abelian Groups is Abelian Group

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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.


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:

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

$\blacksquare$


Also see


Sources