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 $\text G 0$: Closure that all the powers of $x$ are elements of $H$.

That is:

$\gen x \le H$

By Order of Cyclic Group equals Order of Generator:

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

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

Hence the result.

$\blacksquare$

Sources

• 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts