Subset Product of Subgroups/Sufficient Condition
Theorem
Let $\struct {G, \circ}$ be a group.
Let $H, K$ be subgroups of $G$.
Let $H$ and $K$ be permutable subgroups of $G$.
That is, suppose:
- $H \circ K = K \circ H$
where $H \circ K$ denotes subset product.
Then $H \circ K$ is a subgroup of $G$.
Proof 1
Suppose $H \circ K = K \circ H$.
First note that $H \circ K \ne \O$, as $e_G = e_G \circ e_G \in H \circ K$, from Identity of Subgroup.
Suppose $a_1, a_2 \in H, b_1, b_2 \in K$.
Then:
- $\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = a_1 \circ \paren {b_1 \circ a_2} \circ b_2$.
Since $H \circ K = K \circ H$, we see that, for some $a \in H, b \in K$:
- $b_1 \circ a_2 = a \circ b$
Thus:
- $\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = \paren {a_1 \circ a} \circ \paren {b \circ b_2}$
As $H, K \le G$, we have $a_1 \circ a \in H$ and $b \circ b_2 \in K$, hence:
- $\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} \in H \circ K$
thus demonstrating closure of $H \circ K$ under $\circ$.
Finally, if $a \circ b \in H \circ K$, then by Inverse of Group Product:
- $\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$
Since $b^{-1} \in K$ and $a^{-1} \in H$, we have:
- $\paren {a \circ b}^{-1} \in K \circ H$
and hence $H \circ K$ is shown to be closed under inverses.
Thus, from the Two-Step Subgroup Test, $H \circ K$ is a subgroup of $G$.
$\blacksquare$
Proof 2
Suppose $H \circ K = K \circ H$.
Then:
\(\ds \paren {H \circ K} \circ \paren {H \circ K}^{-1}\) | \(=\) | \(\ds H \circ K \circ K^{-1} \circ H^{-1}\) | Inverse of Product of Subsets of Group | |||||||||||
\(\ds \) | \(=\) | \(\ds H \circ K \circ K \circ H\) | Inverse of Subgroup | |||||||||||
\(\ds \) | \(=\) | \(\ds H \circ K \circ H\) | Product of Subgroup with Itself | |||||||||||
\(\ds \) | \(=\) | \(\ds H \circ H \circ K\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds H \circ K\) | Product of Subgroup with Itself |
That is:
- $\paren {H \circ K} \circ \paren {H \circ K}^{-1} = H \circ K$
Thus by definition of set equality:
- $\paren {H \circ K} \circ \paren {H \circ K}^{-1} \subseteq H \circ K$
So from One-Step Subgroup Test using Subset Product:
- $H \circ K \le G$
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): $\S 5$: Proposition $5.17$