# Let $G$ be a group and $S$ be a subset of $G$. Show that $\langle S\rangle=\{ a_1 \dots a_n \mid a_i \in S\textrm{ or }a_i^{-1} \in S;n \geq 1 \}$

Let $G$ be a group and $S$ be a nonempty subset of $G$. Show that $\langle S\rangle=\lbrace a_1 \dots a_n \mid a_i \in S\textrm{ or }a_i^{-1} \in S;n \geq 1 \rbrace$. How to show this? What I can think of is to consider a set $A=\lbrace a_1 \dots a_n \mid a_i \in S\textrm{ or }a_i^{-1} \in S;n \geq 1 \rbrace$ and try to prove that $\langle S\rangle=A$. Anyone can guide me ?

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This is false when $S$ is empty: you need to explicitly include the identity element in this case. –  Chris Eagle Feb 1 '13 at 9:01

$\langle S\rangle$ is the smallest subgroup of $G$ that contains $S$. Let $A=\{a_1\cdots a_n\mid a_i \in S , a^{-1}\in S\}$. To show $\langle S\rangle=A$ the strategy is to prove that $A\subseteq \langle S\rangle$ and that $\langle S\rangle\subseteq A$.

$\langle S\rangle\subseteq A$ will follow immediately from the definition of $\langle S\rangle$ by showing that $A$ is a subgroup of $G$ that contains $A$. That is very easily done.

For the other direction then, as $S\subseteq \langle S\rangle$ (obviously) and since $\langle S\rangle$ is a subgroup it follows that any expression of the form $a_1\cdots a_n$ with, for each $i$, either $a_i\in S$ or $a_i^{-1}\in S$ must be in $\langle S\rangle$ (apply induction on the closure of subgroup under inverses and binary products). This shows that $A\subseteq \langle S\rangle$ and completes the proof.

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for the $A \subset \langle S \rangle$, can I let $a_1 ... a_n \in A$, then since each $a_i \in S \subset \langle S \rangle$ for $1 \leq i \leq n$, by closure , $a_1 ... a_n \in \langle S \rangle$? –  Idonknow Feb 2 '13 at 21:09
for the $\langle S \rangle \subseteq A$ , I don understand why it follows after we show that $A$ is a subgroup of $G$ ? –  Idonknow Feb 2 '13 at 21:11
A is a subgroup of G that contains A. <S> is the smallest subgroup of G that contains A. So <S> is contains in A. –  Ittay Weiss Feb 2 '13 at 22:01
I thought it should be '$\langle S \rangle$ is the smallest subgroup of G that contains $S$' ? –  Idonknow Feb 3 '13 at 3:41
yes, sorry, a typo. <S> is the smallest subgroup of G that contains S. –  Ittay Weiss Feb 3 '13 at 3:42

Besides to what you asked as $\langle S\rangle$ and @Ittay cited in details; we can also show that: $$\langle S\rangle=\{a_1^{\epsilon_1}a_2^{\epsilon_2}\cdot\cdot\cdot a_n^{\epsilon_n}\mid a_i\in S,~~\epsilon_i\in\Bbb Z,~ n=1,2,...\}$$ The proof is routine and is similar to @Ittay's illustration. For example, if $G=\mathbb Z$, then $$\langle 2\rangle=\{2k\mid k\in\mathbb Z\}=2\mathbb Z$$ This is what you asked some things about it in your previous question. Or if $G=\mathbb Q$ be the additive group then $$\langle \frac{1}2, 3\rangle=\{3k+\frac{s}2\mid k,s\in\mathbb Z\}$$ Note that in any cases $S$ shoud be non empty subset of $G$.

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Nice addition :-) + –  amWhy Feb 1 '13 at 14:46