External Direct Product of Groups

Theorem

Let $\left({G_1, \circ_1}\right)$ and $\left({G_2, \circ_2}\right)$ be groups whose identity elements are $e_1$ and $e_2$ respectively.

Let $\left({G_1 \times G_2, \circ}\right)$ be the external direct product of $G_1$ and $G_2$.

Then $\left({G_1 \times G_2, \circ}\right)$ is a group whose identity element is $\left({e_1, e_2}\right)$.

General Result

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

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

Proof

Taking the group axioms in turn:

G0: Closure

Follows from External Direct Product Closure.

$\Box$

G1: Associativity

Follows from External Direct Product Associativity.

$\Box$

G2: Identity

Follows from External Direct Product Identity.

$\Box$

G3: Inverses

Follows from External Direct Product Inverses.

$\Box$

All group axioms are fulfilled, hence the result.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 13$: Theorem $13.1$: Corollary