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

Question

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 ?

Answer 1

$\langle S\rangle$ 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.

Answer 2

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 $G$.