prove that $\langle S \rangle$ is the intersection of all subgroups containing $S$ Let $S$ be a subset of $G$, where $G$ is a group. I have to prove that the generated of $S$ is the intersection of all subgroups which contain $S$. I have been thinking about this for a while, but it seems I cant find a way of proving it. I really need hints! Thank you.

$A = \{H \le G| S \subset H \}$. Note that $G \in A$, therefore $A \neq \phi$ and $S' := \cap_{H \in A} H$ is well defined. We will now prove both the inclusions $S' \subset \langle S\rangle $ and $\langle S\rangle \ \subset S'$.
By definition, the generated by $S$ is the smallest subgroup which contains $S$: this implies that $\langle S\rangle  \in A$ therefore $S' \subset \langle S\rangle $. The other inclusion follows the fact that if we suppose $\langle S\rangle  \nsubseteq S'$ then $\langle S\rangle  \cap \ S'$ contradicts the minimality of $\langle S\rangle $.
