Exercise 14.3.5

Question

\def\imp{\Rightarrow}
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Let $A$ and $B$ be non-Abelian simple groups. You will show that $A 	imes \{e_B\}$ and $\{e_A\}	imes B$ are the only nontrivial normal subgroups of $A 	imes B$. Let $N \subset A 	imes B$ be a normal subgroup different from $\{(e_A,e_B)\},A 	imes\{e_B\}$, and $\{e_A\} 	imes B$.
\be
\item[(a)] Prove that $A 	imes\{e_B\}$ and $\{e_A\}	imes B$ are normal in $A 	imes B$. Hence, if we can show that $N = A	imes B$, then we will be done.
\item[(b)] Prove that we can find $(a,b)\in N$ such that $e_A 
e a \in A$ and $e_B 
e b\in B$.
\item[(c)] Let $(a,b)\in N$ be as in part (b). Show that $(aa_1a^{-1}a_1^{-1}, e_B) \in N$ for any $a_1 \in A$.
\item[(d)] Given $e_A
e a \in A$, prove that there is $a_1\in A$ such that $aa_1
e a_1a$. Then combine this with parts (b) and (c) to show that $N\cap(A	imes\{e_B\}) = A 	imes\{e_B\}$.
\item[(e)] Part (d) implies that $A 	imes \{e_B\} \subset N$, and the inclusion $\{e_A\} 	imes B \subset N$ is proved similarly. Use this to prove that $N = A	imes B$.
\ee
Exercise 18 will explore various aspects of this argument.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

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ewcommand{\ssi} {si et seulement si }

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\begin{proof}
\item[(a)] Take $(a_0,e_B)$ in $A	imes\{e_B\}$, and $(a,b) \in A	imes B$. Then
\begin{align*}
(a,b)(a_0,e_B)(a,b)^{-1} 
&=(a,b)(a_0,e_B)(a^{-1},b^{-1})\
&=(aa_0a^{-1},be_Bb^{-1})\
&=(aa_0a^{-1}, e_B) \in A 	imes\{e_B\},
\end{align*}
thus $A	imes \{e_B\}$ is normal in $A 	imes B$, and similarly $\{e_A\} 	imes B$ is normal in $A 	imes B$.

\item[(b)] Reasoning by contradiction, suppose that we can't find $(a,b) \in N$ such that $a 
e e_A,b
e e_B $ , then $N \subset (A 	imes \{e_B\}) \cup (\{e_A\} 	imes B)$.

Note that $A	imes \{e_B\} \simeq A$ is simple. If $N \subset A 	imes \{e_B\}$, then $N$ is normal in $A 	imes \{e_B\} \subset A 	imes B$, thus $N = A	imes\{e_B\}$ or $N = \{(e_A,e_B)\}$, but this contradicts the assumptions on $N$. Therefore $N 
ot \subset A 	imes \{e_B\}$, and similarly $N 
ot \subset \{e_A\} 	imes B$, so that
\begin{align*}
&N \subset (A 	imes \{e_B\}) \cup (\{e_A\} 	imes B), \
&N 
ot \subset A 	imes \{e_B\}, \quad N 
ot \subset \{e_A\} 	imes B.
\end{align*}
Since $N \subset (A 	imes \{e_B\}) \cup (\{e_A\} 	imes B)$ and $N 
ot \subset A 	imes \{e_B\}$, there exists some $n_1 = (a',b) \in N$ such that $n_1 = (a',b) \in \{e_A\} 	imes B$ and $n_1 = (a',b) 
ot \in A	imes \{e_B\}$. Thus $n_1 = (e_A,b) \in N, b 
e e_B$.

Similarly, since  $N \subset (A 	imes \{e_B\}) \cup (\{e_A\} 	imes B)$ and $N 
ot \subset \{e_A\} 	imes B$, there exists some $n_2 = (a,b') \in N$ such that $n_2 = (a,b') \in A 	imes
 \{e_B\}$ and $n_2 = (a,b') 
ot \in \{e_A\} 	imes B$. Thus $n_2 = (a,e_B) \in N, a 
e e_A$.

Then $n = n_1n_2 =(e_A,b)(a,e_B) = (a,b) \in N$, and $a 
e e_A,b 
e e_B$.

\item[(c)] $N$ is normal, and  $(a^{-1},b^{-1}) = (a,b)^{-1} \in N$, thus
$$(a_1,e_B)(a,b)^{-1}(a_1,e_B)^{-1} = (a_1 a^{-1}a_1^{-1},b^{-1}) \in N.$$
Left multiplication by $(a,b) \in N$ gives
$$(a,b)[(a_1,e_B)(a,b)^{-1}(a_1,e_B)^{-1} ] = (aa_1a^{-1}a_1^{-1}, e_B) \in N.$$

\item[(d)] Let $a
e e_A$ be a fixed element in $A$.

Consider the center of $A$:
$$Z(A) = \{x \in A\, | \, \forall y \in A, xy = yx\}.$$
$Z(A)$ is a normal subgroup of $A$ (it is the kernel of $f : A 	o \mathrm{Aut(A)}$ defined by $f(y) =\phi_y$, where $\phi_y(x) = yxy^{-1}$ for all $x\in A$). Since $A$ is non-Abelian, $Z(A) 
e A$, and since $A$ is simple, $Z(A) = \{e_A\}$.

Therefore, $a
e e_A$ implies $a 
ot \in Z(A)$, thus there exists $a_1 \in A$ such that $aa_1 
e a_1 a$.

Since $N$ and $A	imes \{e_B\}$ are normal subgroups of $A	imes B$, the intersection $N \cap (A	imes \{e_B\})$ is normal in $A	imes B$, a fortiori in the simple group $A	imes \{e_B\}$. Therefore  $N \cap (A	imes \{e_B\}) =\{(e_A,e_B)\}$ or $N \cap (A	imes \{e_B\}) = A 	imes \{e_B\}$. But $N \cap (A	imes \{e_B\}) =\{(e_A,e_B)\}$ is impossible, because $(e_A,e_B) 
e (aa_1a^{-1}a_1^{-1}, e_B) \in N$. Hence $N \cap (A	imes \{e_B\}) = A 	imes \{e_B\}$.

\item[(e)] Part (d) implies that $A 	imes \{e_B\} \subset N$, and the inclusion $\{e_A\} 	imes B \subset N$ is proved similarly. Let $(a,b)$ be any element in $A	imes B$. Then 
$n_1 = (a,e_B) \in A	imes\{e_B\} \subset N$, and $n_2 = (e_A,b) \in \{e_A\} 	imes B \subset N$, thus $n_1 \in N, n_2 \in N$, so that $n =n_1n_2 = (a,b) \in N$. This proves $A	imes B \subset N$, where $N \subset A	imes B$. Therefore $N = A 	imes B$.

To conclude, if $A,B$ are non-Abelian simple groups, the only non trivial normal subgroups of $A	imes B$ are $A 	imes \{e_B\}$ and $\{e_A\} 	imes B$.
\end{proof}

Exercise 14.3.5

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Exercise 14.3.5

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