Product of Subgroup with Itself

Theorem

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

Then:

$\forall H \le G: H \circ H = H$

where:

$H \circ H$ denotes the subset product of $H$ with $H$

$\le$ denotes the subgroup relation.

$H \circ H \subseteq H$

Let $e$ be the identity of $G$.

By Identity of Subgroup, it is also the identity of $H$.

So:

$\ds h$

$\in$

$\ds H$

$\ds \leadsto \ \ $

$\ds e \circ h$

$\in$

$\ds H \circ H$

$\ds \leadsto \ \ $

$\ds h$

$\in$

$\ds H \circ H$

$\ds \leadsto \ \ $

$\ds H$

$\subseteq$

$\ds H \circ H$

Hence the result from the definition of set equality.

$\blacksquare$

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 $9.10 \ \text {(a)}$
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $2$
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 50.1$ Quotient groups