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:

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

Proof

Proof by induction:

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

Basis for the Induction

$P(0)$ is true, as this says:

$\left({g, h}\right)^0 = \left({e_G, e_H}\right)$

$P(1)$ is true, as this 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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({g, h}\right)^{k+1}\)	\(=\)	\(\displaystyle \left({g, h}\right)^k \circ \left({g, h}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({g^k, h^k}\right) \circ \left({g, h}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Induction Hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({g^k \circ_1 g, h^k \circ_2 h}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of external or internal group product
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({g^{k+1}, h^{k+1} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

Therefore:

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

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.

$\blacksquare$