Powers of Elements in Group Direct Product

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

Let $$\left({G \times H, \circ}\right)$$ be the group direct product (either external or internal) of $$G$$ and $$H$$.

Then $$\forall n \in \mathbb{Z}: \forall g \in G, h \in H: \left({g, h}\right)^n = \left({g^n, h^n}\right)$$.

Proof
Proof by induction:

For all $$n \in \mathbb{N}$$, let $$P \left({n}\right)$$ be the proposition $$\forall g \in G, h \in H: \left({g, h}\right)^n = \left({g^n, h^n}\right)$$.


 * $$P(0)$$ is true, as this just says $$\left({g, h}\right)^0 = \left({e_G, e_H}\right)$$.

Basis for the Induction

 * $$P(1)$$ is trivially true, as this just says $$\left({g, h}\right) = \left({g, h}\right)$$. This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 1$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:

$$\left({g, h}\right)^k = \left({g^k, h^k}\right)$$.

Then we need to show:

$$\left({g, h}\right)^{k+1} = \left({g^{k+1}, h^{k+1}}\right)$$.

Induction Step
This is our induction step:

So $$P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall n \in \mathbb{N}: \forall g \in G, h \in H: \left({g, h}\right)^n = \left({g^n, h^n}\right)$$.


 * 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.