# Order of Element in Group equals its Order in Subgroup

## Theorem

Let $G$ be a group.

Let $H\ le G$, where $\le$ denotes the property of being a subgroup.

Let $x \in H$.

Then the order of $x$ in $H$ equals the order of $x$ in $G$.

## Proof

Let $\gen x$ be the subgroup of $G$ generated by $x$.

By definition, $\gen x \le G$.

All the elements of $\gen x$ are powers of $x$.

As $x \in H$ it follows by group axiom $G0$: closure that all the powers of $x$ are elements of $H$.

That is:

$\gen x \le H$
$\order x = \order {\gen x}$

But $\gen x$ is the same cyclic group whether it is considered as embedded in $H$ or $G$.

Hence the result.

$\blacksquare$