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.