Product of Subgroup with Inverse
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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 $9.10 \ \text {(a)}$