Exercise 14.3.18

Proof. (a) Consider the group $G=\mathbb{Z}∕3\mathbb{Z}\times \mathbb{Z}∕3\mathbb{Z}=A\times B$, where $A=B=\mathbb{Z}∕3\mathbb{Z}$ is simple. This is the additive group of the vector space ${𝔽}_3^2$. If $v=(1,2)$ (for instance), the vectorial line

$$N={𝔽}_3\cdot v=\{(0,0),(1,2),(2,1)\}$$

is a subspace of ${𝔽}_3^2$, therefore a subgroup of $G$. Moreover $N$ is normal and not trivial, and $N\ne \{0\}\times B,N\ne A\times \{0\}$. This counterexample shows that Exercise 5 is false if we drop the assumption that $A$ and $B$ are non-Abelian. (b) We will follow the steps of Exercise 5. $\text{∙}$ First, note that the subsets $H_1\times \cdots H_r\subset G$, where $H_i=\{e_{A_i}\}$ or $H_i=A_i$ for all $i=1,\dots r$, are normal subgroups of $G=A_1\times \cdots \times A_r$:

Take $h=(h_1,\dots ,h_r)\in N=H_1\times \cdots H_r$, and $a=(a_1,\dots ,a_r)\in G$, then $\mathit{ah}{a}^{-1}=(a_1{h}_1{a}_1^{-1},\dots ,a_r{h}_r{a}_r^{-1})$. If $H_i=\{e_{A_i}\}$, then $a_i{h}_i{a}_i^{-1}=a_i{a}_i^{-1}=e_{A_i}\in H_i$, and if $H_i=A_i$, then $a_i{h}_i{a}_i^{-1}\in H_i$, thus $\mathit{ah}{a}^{-1}\in N$.

We will call these subgroups standard normal subgroups. $\text{∙}$ To prove that these standard subgroups are the only normal subgroups of $G$, we use induction on $r$. The case $r=2$ is done is Exercise 5.

For simplicity, we write $e$ for the identity of each group $H_i$.

Now suppose that for any $k<r$, and for any $k$-tuple $(A_1,\dots ,A_k)$ of non-Abelian simple groups, the only normal subgroups of $A=A_1\times \cdots \times A_k$ are $H_1\times \cdots H_k$, where $H_i=\{e\}$ or $H_i=A_i$ for all $i=1,\dots k$.

Now let $A_1,\dots ,A_r$ be non-Abelian simple groups, and let $N$ be a normal subgroup of $A_1\times \cdots \times A_r$. $\text{∙}$ Consider first the case where, for some $i\in \{1,\dots ,n\}$,

$$\begin{array}{llll}\hfill N& \subset A_1\times \cdots \times A_{i-1}\times \{e\}\times A_{i+1}\cdots \times A_r& \hfill & \\ \hfill & \simeq A_1\times \cdots \times A_{i-1}\times A_{i+1}\cdots \times A_r,& \hfill & \end{array}$$ where the isomorphism is the map $(a_1,\dots ,a_{i-1},e,a_{i+1},\dots ,a_r)\mapsto (a_1,\dots ,a_{i-1},a_{i+1},\dots ,a_r)$. This isomorphism sends $N$ on a normal subgroup $N^{\prime }$.

$A_1\times A_{i-1}\times A_{i+1}\cdots \times A_r$ is a direct product of $r-1$ simple non-Abelian subgroups, thus we can apply the induction hypothesis: its only normal subgroups are standard subgroups, thus $N^{\prime }$ is standard. This implies, via the isomorphism, that

$$N=H_1\times \cdots H_{i-1}\times \{e\}\times H_{i+1}\times \cdots \times H_r,\text{ where }{H}_j=\{e\}\text{ or }{H}_j=A_j\text{ for all }j.$$

Therefore $N$ is standard in this case. $\text{∙}$ In the remaining case, $N\not\subset A_1\times \cdots \times A_{i-1}\times \{e\}\times A_{i+1}\times \cdots \times A_r$ for all $i=1,\dots ,r$. We want to prove that $N$ is the whole group $G$. $\text{∙}$ First, we prove that there is some $n=(n_1,\dots ,n_r)\in N$ such that $a_i\ne e$ for all $i,1\le i\le r$.

Fix some $i\in \{1,\dots ,r\}$. Since $N\not\subset A_1\times \cdots \times A_{i-1}\times \{e\}\times A_{i+1}\times \cdots \times A_r$, there exists some $n_i=(a_1,\dots ,a_i,\dots ,a_r)\in N$ such that $a_i\ne e$.

Since $A_i$ is non-Abelian and simple, the centers $Z_i$ of $A_i$ are different from $A_i$, and $Z_i$ is normal in the simple group $A_i$, thus $Z_i=\{e\}$. Using $a_i\ne e$, this shows that there is some $c\in A_i$ such that $a_{i}c\ne c{a}_i$. Now take $a=(e,\dots ,e,c,e,\dots ,e)\in G$. Since $N$ is normal, $n_i(a{n}_i^{-1}{a}^{-1})\in N$, and

$$\begin{array}{llll}\hfill n_{i}a{n}_i^{-1}{a}^{-1}& =(a_1,\dots ,a_i,\dots ,a_r)(e,\dots ,c,\dots ,e)(a_1^{-1},\dots ,a_i^{-1},\dots ,a_r^{-1})(e,\dots ,c^{-1},\dots ,e)& \hfill & \\ \hfill & =(e,\dots ,a_{i}c{a}_i^{-1}{c}^{-1},\dots ,e),& \hfill & \end{array}$$

thus $n_{i}a{n}_i^{-1}{a}^{-1}=(e,\dots ,n_i,\dots ,e)\in N$, where $n_i=a_{i}c{a}_i^{-1}{c}^{-1}\ne e$. Since we can find such $n_i\ne e$ for each index $i$,

$$n=(n_1,\dots ,n_r)=(n_1,e,\dots ,e)(e,n_2,\dots ,e)\cdots (e,\dots ,n_r)\in N,\text{ where }{n}_1\ne e,\dots ,n_r\ne e.$$

$\text{∙}$ Write $b=n_r$. $N$ is normal in $G$, thus, for any $a=(a_1,\dots ,a_{r-1},e)\in A_1\times \cdots \times A_r$,

$$\begin{array}{llll}\hfill n(a{n}^{-1}{a}^{-1})& =(n_1,\dots ,n_{r-1},b)(a_1,\dots ,a_{r-1},e)(n_1^{-1},\dots ,n_{r-1}^{-1},b^{-1})(a_1^{-1},\dots ,a_{r-1}^{-1},e)& \hfill & \\ \hfill & =(n_1{a}_1{n}_1^{-1}{a}_1^{-1},\dots ,n_{r-1}{a}_{r-1}{n}_{r-1}^{-1}{a}_{r-1}^{-1},e)\in N& \hfill & \end{array}$$

Since $Z_i=\{e\}$, we can find $(a_1,\dots ,a_{r-1},e)\in A$ such that $n_i{a}_i{n}_i^{-1}{a}_i^{-1}\ne e$ for every $i,1\le i\le r-1$.

Write for simplicity $A=A_1\times \cdots \times A_{r-1},B=A_r$, and $e_A=(e,\dots ,e)$ the identity of $A$.

By the induction hypothesis, $N\cap (A\times \{e\})$ is a direct product of trivial subgroups $H_i\subset A_i$:

$$\begin{array}{cc}N\cap (A\times \{e\})=H_1\times \cdots \times H_{r-1}\times \{e\},\hfill & H_i=\{e\}\text{ or }{H}_i=A_i,i=1,\dots ,r-1,\hfill \\ \hfill \end{array}$$

Since

$$\mathit{na}{n}^{-1}{a}^{-1}=(n_1{a}_1{n}_1^{-1}{a}_1^{-1},\dots ,n_{r-1}{a}_{r-1}{n}_{r-1}^{-1}{a}_{r-1}^{-1},e)\in N\cap (A\times \{e\})=H_1\times \cdots \times H_{r-1}\times \{e\},$$

every $H_i$ contains an element $n_i{a}_i{n}_i^{-1}{a}_i^{-1}\ne e$, thus $H_i=A_i$ for all $i,1\le i\le r-1$. This proves that

$$N\cap (A\times \{e\})=A\times \{e\},$$

thus $A_1\times \cdots \times A_{r-1}\times \{e\}\subset N$.

With the same reasoning, $\{e\}\times A_2\times \cdots \times A_r\subset N.$ $\text{∙}$ If $a=(a_1,\dots ,a_r)$ is any element of $A_1\times \cdots \times A_r$, then

$$a=(a_1,\dots ,a_{r-1},e)(e,\dots ,e,a_r),$$

where $(a_1,\dots ,a_{r-1},e)\in A_1\times \cdots A_{r-1}\times \{e\}\subset N$, and $(e,\dots ,e,a_r)\in \{e\}\times A_2\times \cdots \times A_r\subset N$. Therefore $a\in N$.

This proves $N=A\times B=A_1\times \cdots \times A_r\times A_{r+1}$: for every normal subgroup $N$ of $G$ such that there is some $n=(n_1,\dots ,n_r)\in N$ where $n_i\ne e$ for all $i$, $N$ is the whole group $G$. In this case, $N$ is also standard, and the induction is done.

To conclude, if $A_1,\dots ,A_r$ are non-Abelian simple groups, the only normal subgroups of $A_1\times \cdots \times A_r$ are the subgroups

$$N=H_1\times \cdots \times H_r,\text{ where }{H}_i=\{e_{A_i}\}\text{ or }{H}_i=A_i\text{ for all }i\in \{1,\dots ,r\}.$$

□