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(i) Prove that < S > is a subgroup of G, called the subgroup generated by S.

(ii) Let G denote the dihedral group Dn, ρ a rotation of order n, and σ any reflection. Prove that < {ρ, σ} >= G.

I need some help with this problem. Firstly what exactly does the intersection symbol represent at the beginning of the index notation?

On the second part of the question, I can see intuitively why that is the case for any Dn but I am not entirely sure how to prove it for the general case

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    $\bigcap \{H < G : S \subset H\}$ is "The intersection of all subgroups $H$ of $G$, such that $S$ is a subset of $H$"; that is, the intersection of all subgroups containing the set $S$. So I should point out it's not index notation, it's set-builder notation. – 2017-02-23
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    So does that mean that generates this intersection? If so, then wouldn't =S since the intersection of all subgroups with set S IS S? – 2017-02-23
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    $\langle S \rangle$ is *defined* by this intersection, in your case. But since $S$ is only a sub*set*, not necessarily a sub*group*, we can't say that $\langle S \rangle = S$ (and in fact it won't be, if $S$ isn't a subgroup). Take for example $G = \Bbb Z_{12}$ (the cyclic group with $12$ elements). You should find that here $\langle 3 \rangle$ consists of four elements, not just $3$ itself. – 2017-02-23
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    One more question, is the intersection of subgroups the collection of elements they each have in common? i.e. pretty much the same definition as the intersection of subsets? – 2017-02-23
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    Yes, exactly! It's defined the same way, it's just there are stricter criteria for getting into the "have your intersection taken" club than just being a subset of the group.. – 2017-02-23

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