Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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
up vote 2 down vote accepted

$\langle S\rangle$ 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.

share|cite|improve this answer
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 $G$.

share|cite|improve this answer
Nice addition :-) + – amWhy Feb 1 '13 at 14:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.