Power of Element in Subgroup

Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $\struct {H, \circ}$ be a subgroup of $\struct {G, \circ}$.

Let $x \in H$.

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

Proof
Proof by induction:

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

$\map P 0$ 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
$\map P 1$ 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 $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ 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 $x^{-\paren {k + 1} } \in H$.

So $\map P k \implies \map P {k + 1}$.

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.