Prove that $\langle S \rangle = \bigcap_{S\subseteq H, H\le G}H$

Question

Prove that $\langle S \rangle = \bigcap_{S\subseteq H, H\le G}H$ where $\langle S \rangle = \{x_1, . . . , x_m \mid x_i \in S\cup S^-, m \in \mathbb{N} \}$

I know how to prove that $\langle S \rangle$ is a subgroup of G by using the subgroup criterion but I'm not sure how to approach this question.

I'm guessing I will have to show $\langle S \rangle$ is composed of disjoint subsets of $H$ but I don't know how to begin.

Please can someone give me hint!

Answer 1

Let $K=\displaystyle\bigcap_{S\subseteq H, H\le G}H$.

Clearly $\langle S \rangle \subseteq H$ for all subgroups $H$ containing $S$. This implies that $\langle S \rangle \subseteq K$.

Since $\langle S \rangle$ is a subgroup that clearly contains $S$, one of the $H$'s in $K$ is $\langle S \rangle$, and so $K \subseteq \langle S \rangle$.

You also want to prove this characterization of $\langle S \rangle$:

$\langle S \rangle$ is the smallest subgroup of $G$ that contains $S$, in the sense that if $H$ is a subgroup of $G$ that contains $S$, then $\langle S \rangle \subseteq H$.