Exercise 3.1.23 (Join of a collection of normal subgroups)

Proof. Not as straightforward as Exercise 22.

(a)

We first prove for training that $H\vee K⊴G$ if $H⊴G$ and $K⊴G$.

As usual, we define $\mathit{HK}$ by $\mathit{HK}=\{\mathit{hk}\mid h\in H,k\in K\}$. The subset $\mathit{HK}$ is not always a subgroup of $G$, but we will prove that if $H⊴G$ and $K⊴G$, then $\mathit{HK}$ is a subgroup of $G$, and $H\vee K=\mathit{HK}$.

First $\mathit{HK}$ is a subgroup of $G$: Since $1=1\cdot 1\in \mathit{HK}$, then $\mathit{HK}\ne \varnothing$. If $u\in \mathit{HK}$ and $v\in \mathit{HK}$, then $u=\mathit{hk},v=h^{\prime }{k}^{\prime }$ for some elements $h,h^{\prime }\in H$ and $k,k^{\prime }\in K$. Then

$$\begin{array}{llll}\hfill u{v}^{-1}& =(\mathit{hk}){(h^{\prime }{k}^{\prime })}^{-1}& \hfill & \\ \hfill & =\mathit{hk}{k{}^{\prime }}^{-1}{h{}^{\prime }}^{-1}& \hfill & \\ \hfill & =(h{h{}^{\prime }}^{-1})(h^{\prime }k{k{}^{\prime }}^{-1}{h{}^{\prime }}^{-1})& \hfill & \\ \hfill & =h^{″}{k}^{″},& \hfill & \end{array}$$

where $h^{″}=h{h{}^{\prime }}^{-1}\in H$ and $k^{″}=h^{\prime }(k{k{}^{\prime }}^{-1}){h{}^{\prime }}^{-1}\in K$, because $K⊴G$. Therefore $u{v}^{-1}\in \mathit{HK}$ so $\mathit{HK}$ is a subgroup of $G$. Suppose that $L$ is any subgroup of $G$ such that $L\supseteq H$ and $L\supseteq K$. If $u=\mathit{hk}\in \mathit{HK}$, then $h\in L$ and $k\in L$, thus $u=\mathit{hk}\in L$, so $\mathit{HK}\le L$. This shows that $\mathit{HK}$ is the smallest subgroup of $G$ which contains $H$ and $K$, so

$$H\vee K=\mathit{HK}\le G,(\text{if }H⊴G,K⊴G).$$

Now if $g\in G$ and $u=\mathit{hk}\in \mathit{HK}$ $(h\in H,k\in K)$, since $H⊴G$ and $K⊴G$,

$$\mathit{gu}{g}^{-1}=\mathit{ghk}{g}^{-1}=(\mathit{gh}{g}^{-1})(\mathit{gk}{g}^{-1})\in \mathit{HK}.$$

Therefore

$$H\vee G=\mathit{HK}⊴G.$$

(b)

Suppose that $H_1,H_2,\dots ,H_n$ are normal subgroups of $G$. We prove by induction that $H_1{H}_2\cdots H_n$ is a normal subgroup of $G$.

By part (a), $H_1{H}_2$ is a normal subgroup of $G$. Assume that $H=H_1{H}_2\cdots H_{k-1}$ is a normal subgroup of $G$, for some $k$,  $1\le k\le n$.

Then $H⊴G$, and $H_k⊴G$, therefore, by part (a),

$$H_1{H}_2\cdots H_k=H{H}_k⊴G.$$

The induction is done, which prove that if $H_1,H_2,\dots ,H_n$ are normal subgroups of $G$, then $H_1{H}_2\cdots H_n$ is a normal subgroup of $G$

(c)

We suppose that ${(H_i)}_{i\in I}$ is a family of normal subgroups of $G$, where $I\ne \varnothing$ is any set of indices.

Consider the subset $P$ of $G$ whose elements are the the products ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{a}_i$ where all $a_i=1$, except for finitely many indices $i\in I$. More formally, if ${𝒫}_f(I)$ denotes the set of finite subsets of $I$, then

$$P=\left\{x\in G\mid \exists <!--\; FUNCTION\; APPLICATION\; -->S\in {𝒫}_f(I),\left(x=\underset{i\in S}{\prod }{a}_i\text{and}\forall <!--\; FUNCTION\; APPLICATION\; -->i\in S,a_i\in H_i\right)\right\}.$$

• $P$ is a subgroup of $G$: First $1={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in \varnothing }{a}_i\in P$, so $P\ne \varnothing$.

  Let $x=\underset{i\in S}{\prod <!--\; FUNCTION\; APPLICATION\; -->}{a}_i$ and $y=\underset{i\in S^{\prime }}{\prod <!--\; FUNCTION\; APPLICATION\; -->}{b}_i$ be elements of $P$, where $S$ and $S^{\prime }$ are finite subsets of $I$, and $a_i,b_i\in H_i$ for all $i$.

  Put $T=S\cup S^{\prime }=\{k_1,\dots ,k_n\}\in {𝒫}_f(I)$. Then if we put $a_i=1$ if $i\in T-S$ and $b_i=1$ if $i\in T-S^{\prime }$, we may write

  $$x=\underset{i\in T}{\prod }{a}_i=a_{k_1}{a}_{k_2}\cdots a_{k_n},y=\underset{i\in T}{\prod }{b}_i=b_{k_1}{b}_{k_2}\cdots b_{k_n},$$

  where $a_{k_i}\in H_{k_i}$ and $b_{k_i}\in H_{k_i}$ for all $i=1,2,\dots ,n$. Therefore

  $$x\in H_{k_1}{H}_{k_2}\cdots H_{k_n}\text{and}y\in H_{k_1}{H}_{k_2}\cdots H_{k_n}.$$

  By part (b), $H_{k_1}{H}_{k_2}\cdots H_{k_n}$ is a subgroup of $G$, thus $x{y}^{-1}\in H_{k_1}{H}_{k_2}\cdots H_{k_n}\le P$, so $x{y}^{-1}\in P$. This shows that $P$ is a subgroup of $G$.

• $P$ is the smallest subgroup of $G$ which contains every $H_i$: Indeed $H_i\le P$ for each $i\in I$ and if $L$ is any subgroup of $G$ which contains ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{H}_i$, then every $x={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in S}{a}_i$, where $S$ is a finite subset of $I$ and $a_i\in H_i$ for all $i\in S$, then $x\in L$, so $P\subset L$. This shows that

  $$P=\underset{i\in I}{\vee }{H}_i=⟨\underset{i\in I}{\bigcup }{H}_i⟩$$

  is the smallest subgroup of $G$ which contains every $H_i$.

• $P$ is normal in $G$: Let $x={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in S}{a}_i$ be any element of $P$, where $S$ is a finite subset of $I$ and $a_i\in H_i$ for all $i\in S$. Then

  $$\mathit{gx}{g}^{-1}=\underset{i\in S}{\prod }g{a}_i{g}^{-1}=\underset{i\in S}{\prod }{b}_i,$$

  where $b_i=g{a}_i{g}^{-1}\in H_i$. Therefore $\mathit{gx}{g}^{-1}\in N$.

These three items show that

$$\underset{i\in I}{\vee }{H}_i⊴G.$$