# Cartesian Product of Group Actions

## Theorem

Let $\struct {G, \circ}$ be a group.

Let $S$ and $T$ be sets.

Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be group actions.

Then the operation $*: G \times \paren {S \times T} \to S \times T$ defined as:

$\forall \tuple {g, \tuple {s, t} } \in G \times \paren {S \times T}: g * \tuple {s, t} = \tuple {g *_S s, g *_T t}$

is a group action.

## Proof

The group action axioms are investigated in turn.

Let $g, h \in G$ and $s, t \in S$.

Thus:

 $\displaystyle g * \tuple {h * \tuple {s, t} }$ $=$ $\displaystyle g * \tuple {h *_S s, h *_T t}$ Definition of $*$ $\displaystyle$ $=$ $\displaystyle \tuple {g *_S \tuple {h *_S s}, g *_T \tuple {h *_T t} }$ Definition of $*$ $\displaystyle$ $=$ $\displaystyle \tuple {\paren {g \circ h} *_S s, \paren {g \circ h} *_T t}$ Group Action Axiom $GA \, 1$ for both $*_S$ and $*_T$ $\displaystyle$ $=$ $\displaystyle \paren {g \circ h} * \tuple {s, t}$ Definition of $*$

demonstrating that Group Action Axiom $GA \, 1$ holds.

Then:

 $\displaystyle e * \tuple {s, t}$ $=$ $\displaystyle \tuple {e *_S s, e *_T t}$ Definition of $*$ $\displaystyle$ $=$ $\displaystyle \tuple {s, t}$ Group Action Axiom $GA \, 2$ for both $*_S$ and $*_T$

demonstrating that Group Action Axiom $GA \, 2$ holds.

The group action axioms are thus seen to be fulfilled, and so $*$ is a group action.

$\blacksquare$