If $G_1$ and $G_2$ are simple groups, what can we say about normal subgroups of $G_1 \times G_2$? If $G_1$ and $G_2$ are simple groups, what can we say about normal subgroups of $G_1 \times G_2$? I remember when I was taking Algebra I this was brought up in the class but at the time the professor left it for us to think over it.
Well, I remember I conjectured that $\{e\}, G_1 \times \{e\}, \{e\}\times G_2$ and $G_1 \times G_2$ are the only normal subgroups of $G_1 \times G_2$. I'm trying to see whether I am right or not. I haven't studied Sylow theorems yet and at the time the professor brought up this question I remember that we were studying direct products of groups. Is it possible to answer this question with elementary theorems in abstract algebra? Notice that $G_1$ and $G_2$ are not restricted to finite groups in my question.
 A: Hints: If $N$ is a normal subgroup of $G_1\times G_2$, as the projections $p_i:G_1\times G_2\to G_i$ are surjective, they map $N$ onto a normal subgroup.
For the case both $p_i(N)=G_i$, let $M:=\{g_1\mid (g_1,1)\in N \}$. Then $M$ is again a normal subgroup (of $G_1$). If $M=G_1$ we are ready soon. 
Finally, if $M=\{1\}$ (by symmetry it also means $\{g_2\mid (1,g_2)\in N\}=\{1\}$), conclude that $N$ is a graph of an isomorphism (e.g. $(g_1,g_2)\in N$ and $(g_1,g_2')\in N$ implies $g_2=g_2'$), in other words $N$ is just the diagonal of $G_1\times G_2$ applying the isomorphism $G_1\cong G_2$.
On the other hand, unless $G$ is commutative, the diagonal of $G\times G$ for a simple group is not a normal subgroup.
A: If $G_1$, $G_2$ have order $p$, for some prime $p$, then $G_1\times G_2$ will have $p+3\ge5$ subgroups, all of which are normal since $G_1\times G_2$ is Abelian. In every other case, $G_1\times G_2$ will have exactly have exactly four normal subgroups: $\left\{1\right\}\times\left\{1\right\}$, $G_1\times\left\{1\right\}$, $\left\{1\right\}\times G_2$, and $G_1\times G_2$.
Proof: First note that $\left\{1\right\}\times\left\{1\right\}$, $G_1\times\left\{1\right\}$, $\left\{1\right\}\times G_2$, and $G_1\times G_2$ are all normal subgroups of $G_1\times G_2$. Since simple groups always have at least two elements, these four subgroups are distinct.
Case One: $G_1$, $G_2$ are isomorphic Abelian groups. In this case $G_1$, $G_2$ are finite groups of order $p$, for some prime $p$. Every non-trivial proper subgroup of $G_1\times G_2$ has order $p$, and is therefore cyclic. Every non-identity element of $G_1\times G_2$ generates a cyclic group of order $p$, and every cyclic subgroup of order $p$ contains $p-1$ non-identity elements. Hence, there are $\dfrac{p^2-1}{p-1}=p+1$ subgroups of order $p$, and there are $p+3$ subgroups altogether, all of which are normal since $G_1\times G_2$ is Abelian.
Case Two: $G_1$, $G_2$ are non-isomorphic Abelian groups. In this case $G_1$, $G_2$ are finite groups of order $p_1$, $p_2$, for distinct primes $p_1$, $p_2$. In this case $G_1\times G_2$ is cyclic. Hence, $G_1\times G_2$ has a unique subgroup of order $d$ for each $d$ that divides the order of the group. Hence $G_1\times G_2$ has exactly four subgroups. So the four normal subgroups listed above are the only subgroups of $G_1\times G_2$.
Case Three: One of $G_1$, $G_2$ is not Abelian. Without any loss of generality, we can suppose it's $G_1$.
For $i=1,2$ let $\pi_i:G_1\times G_2\to G_i$ be the projection homomorphism, which send $(x_1,x_2)\mapsto x_i$. For any normal subgroup $N$ of $G_1\times G_2$, it is easy to check that $\pi_i(N)$ is normal in $G_i$. Since each $G_i$ has exactly two normal subgroups, there are four possibilities. If one of $\pi_1(N)$, $\pi_2(N)$ is trivial, then it is easy to show that $N$ is one of $\left\{1\right\}\times\left\{1\right\}$, $G_1\times\left\{1\right\}$, $\left\{1\right\}\times G_2$. So suppose $\pi_i(N)=G_i$ for $i=1,2$. Our goal is to show that in this case $N=G_1\times G_2$.
Let $K=\left\{x\in G_1\,|\,(x,1)\in N\right\}$. It is easy to show that $K$ is normal in $G_1$. So either $K$ is trivial or $K=G_1$. It will follow from the fact that $G_1$ is non-Abelian that $K=G_1$. Pick $g,x\in G_1$ that don't commute, so that $gxg^{-1}x^{-1}$ is non-trivial. Since $x\in G_1=\pi_1(N)$, there is a $y\in G_2$ with $(x,y)\in N$. Since $N$ is normal, if we conjugate $(x,y)$ by $(g,1)$, we get that $(gxg^{-1},y)\in N$. Since $(x,y),(gxg^{-1},y)\in N$, it follows that 
$$(gxg^{-1},y)\cdot(x,y)^{-1}=(gxg^{-1}x^{-1},1)\in N.$$
Hence $gxg^{-1}x^{-1}$ is a non-trivial element of $K$, so that $K=G_1$.
Now let $(x,y)\in G_1\times G_2$. We want to show that $(x,y)\in N$. Since $y\in G_2=\pi_2(N)$, there is a $a\in G_1$ with $(a,y)\in N$. Also, since $a\in G_1=K$, $(a,1)\in N$. So $(a,y)\cdot(a,1)^{-1}=(1,y)\in N$. Also $x\in G_1=K$, so that $(x,1)\in N$. Hence $(x,1)\cdot(1,y)=(x,y)\in N$. Hence $N=G_1\times G_2$. $\square$
