Powers of Elements in Group Direct Product

Theorem
Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be group whose identities are $e_G$ and $e_H$.

Let $\struct {G \times H, \circ}$ be the group direct product of $G$ and $H$.

Then:
 * $\forall n \in \Z: \forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$

Proof
Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition $\forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$.

Basis for the Induction
$\map P 0$ is true, as this says:
 * $\tuple {g, h}^0 = \tuple {e_G, e_H}$

$\map P 1$ is true, as this says:
 * $\tuple {g, h} = \tuple {g, h}$

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:


 * $\tuple {g, h}^k = \tuple {g^k, h^k}$

Then we need to show:


 * $\tuple {g, h}^{k + 1} = \tuple {g^{k + 1}, h^{k + 1} }$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N: \forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$

So we have shown the result holds true for all $n \ge 0$.

The result for $n < 0$ follows directly from Powers of Group Elements for Negative Indices.