External Direct Product of Groups is Group/Finite Product

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Theorem

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


Proof

Let $\left({G_1, \circ_1}\right), \left({G_2, \circ_2}\right), \ldots, \left({G_n, \circ_n}\right)$ be groups.

Let $\displaystyle \left({G, \circ}\right) = \prod_{k \mathop = 1}^n G_k$ be the external direct product of $\left({G_1, \circ_1}\right), \left({G_2, \circ_2}\right), \ldots, \left({G_n, \circ_n}\right)$.


Taking the group axioms in turn:


G0: Closure

From External Direct Product Closure: General Result it follows that $\left({G, \circ}\right)$ is closed.

$\Box$


G1: Associativity

From External Direct Product Associativity: General Result it follows that $\left({G, \circ}\right)$ is associative.

$\Box$


G2: Identity

Let $e_1, e_2, \ldots, e_n$ be the identity elements of $\left({G_1, \circ_1}\right), \left({G_2, \circ_2}\right), \ldots, \left({G_n, \circ_n}\right)$ respectively.

From External Direct Product Identity: General Result it follows that $\left({e_1, e_2, \ldots, e_n}\right)$ is the identity element of $\left({G, \circ}\right)$.

$\Box$


G3: Inverses

Let $g_1, g_2, \ldots, g_n$ be arbitrary elements of $G_1, G_2, \ldots, G_n$


Let $g_1^{-1}, g_2^{-1}, \ldots, g_n^{-1}$ be the inverse elements of $g_1, g_2, \ldots, g_n$ in $\left({G_1, \circ_1}\right), \left({G_2, \circ_2}\right), \ldots, \left({G_n, \circ_n}\right)$ respectively.


From External Direct Product Inverses: General Result it follows that $\left({g_1^{-1}, g_2^{-1}, \ldots, g_n^{-1}}\right)$ is the inverse element of $\left({g_1, g_2, \ldots, g_n}\right)$ in $\left({G, \circ}\right)$.

$\Box$


All group axioms are fulfilled, hence $\left({G, \circ}\right)$ is a group.

$\blacksquare$


Sources