Subgroup generated by a set

Question

A subgroup generated by a set is defined as (from Wikipedia):

More generally, if S is a subset of a group G, then , the subgroup generated by S, is the smallest subgroup of G containing every element of S, meaning the intersection over all subgroups containing the elements of S; equivalently, is the subgroup of all elements of G that can be expressed as the finite product of elements in S and their inverses.

How does one prove that those statements are equivalent? If the answer is to broad to be presented here, I would appreciate pointers to relevant pages (or books).

Thanks in advance!

Answer 1

This is a typical top-down vs. bottom-up construction of a substructure. See the general discussion here.

Let $S\subseteq G$. We let $$K = \bigcap_{S\subseteq M\leq G}M$$ and $$H = \Bigl\{s_1^{\epsilon_1}\cdots s_m^{\epsilon_m}\mid m\geq 0,\ s_i\in S,\ \epsilon_i\in\{1,-1\}\Bigr\}.$$

We want to show that $K=H$.

Note that if $M$ is a subgroup of $G$, and $S\subseteq M$, then every element of $H$ must be in $M$, since $M$ is closed under products and inverses and contains every $s_i\in S$. Thus, $H\subseteq K$.

Conversely, to prove $K\subseteq H$, it suffices to show that $H$ is a subgroup of $G$ that contains $S$. To see that $S\subseteq H$, let $s\in S$. Letting $m=1$, $\epsilon_1=1$, and $s_1=s$ we have $s\in H$, so $S\subseteq H$.

To see that $H$ is a subgroup of $G$, note that $H$ is nonempty: selecting $m=0$ we obtain the empty product, which by definition is the identity of $G$. So $1\in H$.

Let $s_1^{\epsilon_1}\cdots s_m^{\epsilon_m}$ and $t_1^{\eta_1}\cdots t_n^{\eta_n}$, with $m,n\geq 0$, $\epsilon_i,\eta_j\in\{0,1\}$, and $s_i,t_j\in S$ be elements of $S$. Then $$\Bigl( s_1^{\epsilon_1}\cdots s_m^{\epsilon_m}\Bigr)\Bigl(t_1^{\eta_1}\cdots t_n^{\eta_n}\Bigr)^{-1} = r_1^{\chi_1}\cdots r_{n+m}^{\chi_{n+m}}$$ where $$\begin{align*} r_i &= \left\{\begin{array}{ll} s_i &\text{if }1\leq i\leq m\\ t_{n+m-i+1} & \text{if }m\lt i\leq n+m \end{array}\right.\\ \chi_i &= \left\{\begin{array}{ll} \epsilon_i &\text{if }1\leq i\leq m\\ -\eta_{n+m-i+1} & \text{if }m\lt i\leq n+m \end{array}\right. \end{align*}$$ Note that $r_i\in S$ for each $i$, and $\chi_i\in\{1,-1\}$ for each $i$, so $r_1^{\chi_1}\cdots r_{n+m}^{\chi_{n+m}}$ is an element of $H$. Thus, $H$ is a subgroup of $G$ that contains $S$, and so is one of the subgroups being intersected in the definition of $K$. Hence, $K\subseteq H$.

Since we already had $H\subseteq K$, it follows that $H=K$, as desired.

Answer 2

This is covered in pretty much every introductory group theory text.

Let $H$ be the subset of all elements of $G$ that can be expressed as the finite product of elements in $S$ and their inverses, and let $K$ be the intersection over all subgroups containing the elements of S.

Firstly, note that $H$ is a subgroup of $G$, as the product of two words $U(S)$ and $V(S)$ of finite length gives another word of finite length, $W(S)=U(S)V(S)$, while if $U(S)\in H$ then $U^{-1}(S)$ by a simple inductive argument (the induction step is simply $(V(S)x)^{-1}=x^{-1}V(S)$).

Clearly $S\subseteq H$, so $K\leq H$ (because $K$ is got by intersecting all subgroups containing $S$). On the other hand, $S\subseteq K$ and $K$ must contain all elements of $G$ that can be expressed as the finite product of elements in $K$ and their inverses (because $K$ is closed under (finite) products and inverses), and so $H\leq K$.

Thus, $H=K$ and we're done.

Answer 3

My proof don’t contain anything new compared to already existing ones. My proof is slight variation of existing proof. I saw below definition and theorem from Hungerford’s algebra book.

Definition: Let $G$ be a group and $X\subseteq G$. Let $\{H_i\mid i\in I\}$ be the family of all subgroups of $G$ which contain $X$. Then $\bigcap_{i\in I}H_i$ is called a subgroup of $G$ generated by $X$ and denoted $\langle X\rangle$.

Theorem: If $G$ is a group and $X$ is non empty subset of $G$, then the subgroup $\langle X\rangle$ generated by $X$ consists of all finite products $a_1^{n_1}a_2^{n_2}\cdots a_t^{n_t}$ ($a_i \in X$; $n_i \in \Bbb{Z}$). That is $$\langle X\rangle =\{a_1^{n_1}\cdots a_t^{n_t}\mid a_i\in X\text{ and }n_i\in \Bbb{Z}\}.$$

Proof: $(1)$ Show $H= \{a_1^{n_1}\cdots a_t^{n_t}\mid a_i\in X\text{ and }n_i\in \Bbb{Z}\}$ is a subgroup of $G$ containing $X$. $(2)$ Show if $J\leq G$ and $X\subseteq J$, then $H\subseteq J$.

Let $a_1^{n_1}\cdots a_t^{n_t}, b_1^{m_1}\cdots b_k^{m_k}\in H$. Then $a_1^{n_1}\cdots a_t^{n_t}b_1^{m_1}\cdots b_k^{m_k}\in H$. So $H$ is closed under product $\cdot$ in $G$. Associative law follows from $G$. Since $X\neq \emptyset$, $\exists a\in X$. So $a^0=e\in H$. Let $a_1^{n_1}\cdots a_t^{n_t}\in H$. Then $(a_1^{n_1}\cdots a_t^{n_t})^{-1}=a_t^{-n_t}\cdots a_1^{-n_1}\in H$. Thus $H\leq G$. Let $x\in X$. Then $x^1=x\in H$. So $X\subseteq H$. Hence $X\subseteq H\leq G$.

Suppose $J\leq G$ and $X\subseteq J$. Let $a_1^{n_1}\cdots a_t^{n_t}\in H$, where $a_i\in X$ and $n_i\in \Bbb{Z}$. Since $J\leq G$, in particular $J$ is closed under product in $G$, we have $a_i^{n_i}\in J$ forall $i$. So $a_1^{n_1}\cdots a_t^{n_t}\in J$. Thus $H\subseteq J$. Hence $H$ is the smallest subgroup of $G$ containing $X$, i.e. $H=\langle X\rangle$.