# Power of Element in Subgroup

## Contents

## 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.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 5.2$. Subgroups: Example $89$