# Prove that $\langle S \rangle$ $\leq$ G

Let $G$ be a group and $S \subset G$ where $S \neq \emptyset$. Prove that $\langle S \rangle$ $\leq$ G. In this problem $\langle S \rangle$ is the set generated by S.

Now, I know that if $s \in\langle S \rangle$, then $s=s_1*s_2*s_3 ...*s_n$ where each of the $s_i \in S$. Then because $S \subset G$, and $G$ is a group we have that $s \in G$.

I'm supposed to use the 1-step subgroup test. Showing that if $a,b \in\langle S \rangle$, then $ab^{-1}$ $\in\langle S \rangle$. But i'm stuck.

• Do you mean that $\langle S\rangle$ is the subgroup of $G$ generated by $S$? – SamM Sep 11 '16 at 19:13
• I think so, yes. – 1233211 Sep 11 '16 at 19:15
• Your definition of $S$ doesn't seem to be the right one to me, except if you are asking for something like this. What is precisely your definition of $\langle S \rangle$ ? – Watson Sep 11 '16 at 19:16
• In this case, $\langle S\rangle$ should be defined as the intersection of all subgroups of $G$ containing $S$; intersections of subgroups are again subgroups (easy exercise) and so $\langle S\rangle$ is (trivially) a subgroup. – SamM Sep 11 '16 at 19:18
• It looks as though you are unsure of the definition of $\langle S \rangle$. You cannot hope to solve a problem if you do not know the definitions of the terms involved. – Derek Holt Sep 11 '16 at 19:50

If $a = s_1 s_2 \cdots s_n$ and $b = s_{n+1} s_{n+2} \cdots s_{n+m}$ then $ab^{-1} = s_1 s_2 \cdots s_n s_{n+m}^{-1} \cdots s_{n+2}^{-1} s_{n+1}^{-1}$, which is a product of elements of $S$.
• These should be included in the definition of $\langle S \rangle$. Otherwise the result is false, as can be seen by taking $S = \{1\}$ as subset of the additive group $\mathbb{Z}$. – user133281 Sep 11 '16 at 19:36