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

Question

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

I have answered the question but I believe it's wrong, I wanna know why this answer is not right

Here is my solution:

Let $L \lhd G$ then $L=H_{1}\times K_{1}$ where $H_{1} \leqslant H$ and $K_{1}\leqslant K$

Let $g=(h,k)\in G$ then $L^{g} =L$

$g(H_{1}\times K_{1})g^{-1}=H_{1}^{h}\times K_{1}^{k} = H_{1}\times K_{1}$ then $H_{1}^h=H $ and $ K_{1}^k=K$ for every $h\in H, k\in K$

then $L$ is a product of normal subgroups of $H$ and $K$, since they are simple groups then $L$ would be one of the following, $$\{ G, 1 , H\times 1, 1\times K\}$$

I would appreciate it if you tell where is the mistake?

Answer 1

As @David Craven notes, the error in your solution is that you have no warrant for assuming that a normal subgroup of a direct product is necessarily a product of normal subgroups of the factors. All such subgroups are normal, but in principle there may be other normal subgroups.

For example, your argument never uses that $H$ and $K$ are nonabelian, so if it were correct, you would conclude that the only normal subgroups of $C_p\times C_p$ are $\{e\}\times\{e\}$, $C_p\times\{e\}$, $\{e\}\times C_p$, and $C_p\times C_p$; but this misses at least the "diagonal subgroup" $\{(x,x)\mid x\in C_p\}$, which is also normal (this being an abelian group).

So what you've shown is that the group has at least four normal subgroups, and that any normal subgroup of the given form must be one of those. But you also must prove that every normal subgroup is of this form. And for that, you must use that your groups are nonabelian and simple.

Let $N\triangleleft H\times K$. Then $\pi_H(N)$ is normal in $H$, so either $\pi_H(N)=\{e\}$ or $\pi_H(N)=H$. If $\pi_H(N)=\{e\}$, then $N\leq \{e\}\times K$, and you can proceed as you did. Symmetrically, if $\pi_K(N)=\{e\}$, then you are done.

So assume that $\pi_H(N)=H$ and $\pi_K(N)=K$ (so that $N$ is a subdirect product). I claim that in this case, $N=H\times K$. To prove this, we prove that $(h,e)\in N$ and $(e,k)\in N$ for every $h\in H$, $k\in K$.

But since $H$ is nonabelian and simple, $[H,H]=H$, so it is enough to show that for every $h,h'\in H$, we have $([h,h'],e)\in N$. To that end, let $h,h'\in H$. Since $\pi_H(N)=H$, then there exists $k\in K$ such that $(h,k)\in N$. Conjugating by $(h',e)$, we have that $(h',e)^{-1}(h,k)(h',e) = (h'^{-1}hh',k)\in N$, and therefore, multiplying on the left by $(h,k)^{-1}$, we conclude that $$(h,k)^{-1}(h',e)^{-1}(h,k)(h',e) = (h^{-1}h'^{-1}hh',e) = ([h,h'],e)\in N.$$ Thus, $H\times\{e\}=[H,H]\times\{e\}\leq N$. Symmetrically, $\{e\}\times K\leq N$; thus, $N=H\times K$, and now we are done.

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One could reduce this somewhat by invoking Goursat's Lemma, once you know that you can restrict yourself to subdirect products. Every subdirect product of $H\times K$ corresponds to the following data:

1. A normal subgroup of $H$, $N_1\triangleleft H$;
2. A normal subgroup of $K$, $N_2\triangleleft K$;
3. An isomorphism $\varphi\colon H/N_1\to K/N_2$.

The data gives the "graph subgroup", $$D_{\varphi} = \{(h,k)\in H\times K\mid \varphi(hN_1)=kN_2\}.$$ Since $H$ and $K$ are simple, each of $N_1$ and $N_2$ are either trivial or the whole thing; you cannot have one trivial and the other be the whole thing, because then you do not get the isomorphism in 3. So this only leaves the case where $H=N_1$ and $K=N_2$ (in which case you get $H\times K$), or the case where $N_1=\{e\}$, $N_2=\{e\}$, and $H\cong K$. Then your subgroup is of the form $\{(h,\varphi(h))\mid h\in H\}$ for some isomorphism $\varphi\colon H\to K$, and it is now easy to use the noncommutativity to show that such a subgroup is not normal (I'll leave that to you).

Note that in the commutative case this gives you all the (normal) subgroups of $C_p\times C_p$: in addition to the four "obvious" ones, you also get one for each element of $\mathrm{Aut}(C_p)$, which means there are $p+3$ (normal) subgroups. When $p=2$ you get the Klein $4$-group with its five subgroups.

This is essentially the argument in this question.