Power of Element in Subgroup

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $\left({H, \circ}\right)$ be a subgroup of $\left({G, \circ}\right)$.

Let $x \in H$.

Then:
 * $\forall n \in \Z: x^n \in H$

Proof
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the compound proposition:
 * $\forall n \in \N: x^n \in H$.
 * $\forall n \in \N: x^{-n} \in H$.

$P(0)$ is true, as this just says $x^0 \in H$.

By Powers of Group Elements, $x^0 = e$.

This follows by Identity of Subgroup‎.

Basis for the Induction
$P(1)$ is true, as this just says $x^1 = x \in H$.

By the Two-Step Subgroup Test, we also have that $x^{-1} \in H$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $x^k \in H$.

By the Two-Step Subgroup Test, it follows that $x^{-k} \in H$.

Then we need to show:
 * $x^{k+1} \in H$.

Induction Step
This is our induction step:


 * By Powers of Group Elements, $x^{k+1} = x \circ x^k$.


 * By the base case, $x \in H$.


 * By the induction hypothesis, $x^k \in H$.


 * By the closure axiom, $x \circ x^k \in H$.


 * By the Two-Step Subgroup Test, it follows that $x^{-\left({k+1}\right)} \in H$.

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

Therefore:
 * $\forall n \in \N: x^n \in H$.
 * $\forall n \in \N: x^{-n} \in H$.

Hence the result.