# Condition such that $\langle{S}\rangle=S$

So let $S\subseteq{G}$ where $G$ is a group and $S$ an arbitrary subset. Let $\langle{S}\rangle$ be the subgroup in $G$ generated by $S$. What is the condition such that $\langle{S}\rangle=S$?

My thoughts on this; I'm thinking that $S$ must be a subgroup, or what I mean is that all elements of $S$ must themselves form a subgroup under the binary operation which defines the group. It is trivial that $\langle{G}\rangle=G$. If there is an element $s\in{S}$, then $s^{-1}\in{S}$. The identity must be in $S$, since $\langle{S}\rangle\le{G}$. Since $\langle{S}\rangle$ is closed, generating elements from $S$ would mean that the elements generated would be in $S$. Is this the correct intution?

• $\langle S\rangle$ is by definition a subgroup, and conversely the equality holds true if $S$ is a subgroup. So yes, the equality is logically equivalent to $S$ being a subgroup. – anon Nov 5 '13 at 23:47
• so when it asks for the necessary and sufficient condition, that is it? I though so, helps to have verification. Thanks! – Eleven-Eleven Nov 5 '13 at 23:59

Yes, if $\langle S\rangle = S$, then $S$ is a subgroup because $\langle S\rangle$ is (by definition). Conversely, if $S$ is a subgroup, then the smallest subgroup containing $S$ (which is another way of seeing $\langle S\rangle$) is $S$.
Another way of checking is by taking any two elements $a,b$ in $S$ and verifying that $a*b^{-1}$ is in S, where $*$ is the group operation. This is sufficient for $S$ to be a subgroup.