Subset Product of Subgroups/Necessary Condition/Proof 2

Theorem

Let $\left({G, \circ}\right)$ be a group.

Let $H, K$ be subgroups of $G$.

Let $H \circ K$ be a subgroup of $G$.

Then $H$ and $K$ are permutable.

That is:

$H \circ K = K \circ H$

where $H \circ K$ denotes subset product.

Proof

Suppose $H \circ K$ is a subgroup of $G$.

Then:

\(\ds H \circ K\)

\(=\)

\(\ds \paren {H \circ K}^{-1}\)

Inverse of Subgroup

\(\ds \)

\(=\)

\(\ds K^{-1} \circ H^{-1}\)

Inverse of Product of Subsets of Group

\(\ds \)

\(=\)

\(\ds K \circ H\)

Inverse of Subgroup

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $6$