External Direct Product of Groups is Group

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 $\left({e_1, e_2}\right)$.

Proof
Taking the group axioms in turn:

G0: Closure
From External Direct Product Closure it follows that $\left({G_1 \times G_2, \circ}\right)$ is closed.

G1: Associativity
From External Direct Product Associativity it follows that $\left({G_1 \times G_2, \circ}\right)$ is associative.

G2: Identity
From External Direct Product Identity it follows that $\left({e_1, e_2}\right)$ is the identity element of $\left({G_1 \times G_2, \circ}\right)$.

G3: Inverses
From External Direct Product Inverses it follows that $\left({g_1^{-1}, g_2^{-1}}\right)$ is the inverse element of $\left({g_1, g_2}\right)$ in $\left({G_1 \times G_2, \circ}\right)$.

All group axioms are fulfilled, hence the result.