Subgroup generated by a set

A subgroup created by a set is specified as () :

More usually, if S is a part of a team G, after that, the subgroup created by S, is the tiniest subgroup of G having every component of S, suggesting the junction over all subgroups having the components of S ; equivalently, is the subgroup of all components of G that can be shared as the limited item of components in S and also their inverses.

Just how does one confirm that those declarations are equal? If the solution is to wide to be offered below, I would certainly value reminders to pertinent web pages (or publications).

Many thanks beforehand!

22
2022-07-25 20:43:15
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Answers: 1

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

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.

29
2022-07-25 21:40:27
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