Let G be the direct product of two nonabelian simple groups. Then G has exactly four normal subgroups.

Question

I realise that this question has been posted before here, but while trying to solve it myself, I ran into a block and I'm not sure how to proceed. This is how the question has been laid out in a practice paper:

Question 8 [20 marks] Let $X$ and $Y$ be non-abelian simple groups. Let $G=X\times Y$.
(a) ([10 marks]) Let $N$ be normal subgroup of $G$ not contained in $X\times\{e\}$. Prove that $N$ contains an element of the form $(e,y)$ for some $e\neq y\in Y$.
(b) ([10 marks]) Use item (1) to prove that the only normal subgroupsof $G$ are $G$, $X\times\{e\}$, $\{e\}\times Y$, and the trivial group.

$\quad\:$Remark: You can use item (a) freely for item (b) even if you can not prove item (a).

Here is what I have so far: If $N$ is not contained in $X \times \{e\}$, then it must contain an element of the form $(x, y)$, where $y \in Y-\{e\}$. Since $N$ is normal, taking any element $(a, b) \in G$ and conjugating $(x, y)$ by it, we obtain $(axa^{-1}, byb^{-1}) \in N$. But if $byb^{-1} = e$, then $y=e$, so $byb^{-1} \in Y - \{e\}$. This implies $Y - \{e\}$ is normal, but since $Y$ is simple, $Y - \{e\} = \{e\} \text{ or } Y$, both of which are not possible. This is where I get stuck.

Would also appreciate any hints for part (b), thank you so much!

Answer 1

(a) Consider $\pi_X(N) := \{x \in X: \text{exists $y \in Y$ s.t. $(x, y) \in N$}\} \subseteq X$. This is a normal subgroup of $X$; since $X$ is simple, it is either $\{e\}$ or $X$. If $\pi_X(N) = \{e\}$, then the element of $N$ not in $X \times \{e\}$ must be of the form $(e, y) \in N$ where $y \ne e$. So it remains to handle the case $\pi_X(N) = X$.

Hint: Consider any $x_1, x_2 \in X$. Show that $$(x_1 x_2 x_1^{-1} x_2^{-1}, e) \in N. \tag{$*$}$$

Since $\pi_X(N) = X$, there exists $y \in Y$ such that $(x_2, y) \in N$. Conjugating $(x_2, y) \in N$ by $(x_1, e) \in G$ yields $$(x_1 x_2 x_1^{-1}, y) \in N.$$ Multiplying on the right by $(x_2, y)^{-1} \in N$ yields ($*$).

The commutator subgroup $[X, X] := \langle\{x_1 x_2 x_1^{-1} x_2^{-1} : x_1, x_2 \in X\}\rangle \subseteq X$ is a normal subgroup of $X$, so it is either $\{e\}$ or $X$. Because $X$ is nonabelian, we must have $[X, X] = X$. Since $[X,X] = X$, the result ($*$) implies $(x, e) \in N$ for any $x \in X$.

Finally, consider $(x, y) \in N$ where $y \ne e$, and multiply it by $(x^{-1}, e)$ to obtain $(e, y) \in N$.

Since the last step can be done for any $y \in \pi_Y(N)$, we have actually shown the stronger statement that $(e, y) \in N$ for all $y \in \pi_Y(N)$.

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(b) If $N$ is either contained in $X \times \{e\}$ or contained in $\{e\} \times Y$, then I think you know how to show that it must be $X \times \{e\}$ or $\{e\} \times Y$ or the trivial subgroup.

The remaining case is where $N$ is neither contained in $X \times \{e\}$ nor contained in $\{e\} \times Y$. We know $\pi_X(N)$ and $\pi_Y(N)$ are nontrivial normal subgroups of $X$ and $Y$ respectively, so $\pi_X(N)=X$ and $\pi_Y(N)=Y$. The strengthened version of part (a) implies $N$ contains all elements of the form $(e, y)$ for all $y \in Y$, and similarly all elements $(x, e)$ for all $x \in X$.

I'm not sure how to prove (b) using only (a) as written.