Inverses in Subgroup

Theorem

Let $G$ be a group.

Let $H$ be a subgroup of $G$.

Then for each $h \in H$, the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$.

Proof

Let $h \in H$.

Let:

$h'$ be the inverse of $h$ in $H$

$h^{-1}$ be the inverse of $h$ in $G$.

From Identity of Subgroup:

$h \circ h' = e$

From Inverse in Group is Unique, it follows that $h' = h^{-1}$.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 36.2 \ \text{(ii)}$: Subgroups