Problem: Let $S$ be a subset of a group and let $H$ be the intersection of all subgroups of $G$ that contains $S$. Prove that $\langle S \rangle=H$.
My thoughts: As my goal is to prove $\langle S \rangle=H$ and both are sets, so I can reach my goal by showing $\langle S \rangle \subseteq H$ and $H \subseteq \langle S \rangle$.
My attempt: Let $A:= \{K ≤G| S \subseteq K\}$ and $H:= ∩_{K \in A}K$. Since $S \subseteq K$ for all $K \in A$, thus $S \subseteq H$. This shows $\langle S \rangle \subseteq H$. Now we want to show $H \subseteq \langle S \rangle$ and here i stuck!
My confusion:
I got a solution prove that $\langle S \rangle$ is the intersection of all subgroups containing $S$ . Here I don't understand that if $\langle S \rangle \in A$, then how $H \subseteq \langle S \rangle$?
Best Answer
Hint: To show $\langle S \rangle \subseteq H$, you need to show that $H$ is a subgroup of $G$ containing $S$. To do so, show that the intersection of subgroups is again a subgroup. To see that $H \subseteq \langle S \rangle$, note that $H$ is the intersection of all subgroups of $G$ containing $S$, and that $\langle S \rangle$ is one such subgroup.
Edit: You write in your comment to amrsa that you have shown that the intersection of all subgroups of $G$ containing $S$ contains S. This is true, but not what you need here. Show that for an arbitrary collection $\{K_i\}$ of subgroups of $G$, their intersection $\bigcap_{i \in I} K_i$ is a subgroup of $G$.
For $H \subseteq \langle S \rangle$, if you sit with the definition of $H$ for a bit, you should see that $H \subseteq \langle S \rangle$ is immediate. Think of the case where we have two sets, it's certainly true that $$ A \cap B \subseteq A \text{ and } A \cap B \subseteq B. $$ If $H$ is the intersection of all subgroups containing $S$ and $\langle S \rangle$ is one such subgroup, then it must be that $H \subseteq \langle S \rangle$.
If you want the proof completely in your notation. Let $A = \{K \leq G \mid S \subseteq K\}$, then $H = \cap_{K \in A} K$. As amrsa mentioned, notice that $\langle S \rangle \in A$ and as such $H \subseteq \langle S \rangle$ is immediate because $\langle S \rangle$ is an element of the intersection defining $H$.