Product of Subgroup with Inverse

Theorem

Let $\struct {G, \circ}$ be a group.

Then:

$\forall H \le \struct {G, \circ}:$

$H^{-1} \circ H = H$

$H \circ H^{-1} = H$

where $H \le G$ denotes that $H$ is a subgroup of $G$.

Proof

From Inverse of Subgroup:

$H = H^{-1}$

From Product of Subgroup with Itself:

$H \circ H = H$

The result follows.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.10 \ \text {(a)}$