# 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.

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How have you had the subgroup generated by $S$ defined? – Tobias Kildetoft Mar 20 '13 at 17:52
@TobiasKildetoft As how wikipedia does: en.wikipedia.org/wiki/Generating_set_of_a_group – Badshah Mar 20 '13 at 18:02
In that article it says "intersection of all subgroups containing $S$" among other things. Which part do you mean? – Tobias Kildetoft Mar 20 '13 at 18:05
@TobiasKildetoft I actually mean: <S> is the set wich contains all finite products of elements in S under the group operation. – Badshah Mar 20 '13 at 18:09
Ok, what have you tried? Have you shown if $H$ is a subgroup containing $S$ then $H$ contains the subgroup generated by $S$? – Tobias Kildetoft Mar 20 '13 at 18:11

$A = \{H \le G| S \subset H \}$. 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 $\langle S\rangle$.