# Direct product of simple groups

Let $G=H_1\times H_2$, $H_1,H_2$ are simple groups. Let $L\vartriangleleft G$ ($L$ isn't trivial). Show that $L$ isomorphic to $H_1$ or $H_2$.

I tried to construct "projections" of $L$ on $H_1,H_2$, I only showed that their images must be trivial, but I'm not sure if it's the right direction.

• Hint: What is $L\cap (H_1\times 1)$? Jan 20, 2015 at 14:10
• $H_1\times 1$ or $1\times 1$.. Tried this direction too Jan 20, 2015 at 14:11
• Ok and what is $L\cap (1\times H_2)$? Jan 20, 2015 at 14:12
• can't see why.. what is the problem with $(a,b)\in L$, $a,b\neq 1$ ? Jan 20, 2015 at 14:36
• @j.p. Good point! I've deleted my comment. But I think the proof will need to treat the abelian and mnonabelian cases seperately. Jan 20, 2015 at 15:38

As the projection $p_1 : G \to H_1$ is onto, the image of a normal subgroup of $G$ is normal in $H_1$. As $H_1$ is simple we get that either $p_1(L) = 1$ or $p_1(L) = H_1$.

If $p_1(L) = 1$ then $L$ is a subgroup of $H_2$. As it is a non-trivial normal subgroup of the simple group $H_2$, we get $L=H_2$. (I do not distinguish between inner and outer direct products.)

If $p_1(L) = H_1$ look at the kernel $K = H_2\cap L$ of $p_1$ restricted to $L$. As $K$ is normal in the simple group $H_2$ we know that either $K=1$ or $K=H_2$ holds. In case of $K=1$ we have that $L$ is isomorphic to $H_1$. If $K=H_2$ then it is easy to show that $L=H_1\times H_2 = G$ contradicting that $L$ is a proper subgroup of $G$.

(Also to my surprise no need for extra treatment of the case $H_1 = H_2$ abelian.)

• Exchanging $H_1$ and $H_2$ one sees that the case $H_1 \cap L = 1 = H_2 \cap L$ can only occur when $H_1$ and $H_2$ are isomorphic. Myself's answer shows that both have to be additionally abelian.
– j.p.
Jan 21, 2015 at 14:00

Let me focus on the case where $L\cap (1\times H_1) = L\cap (H_2\times 1) =1$ since from the comments it seems that's where your main difficulty lies. As Derek Holt notices, this case requires some special attention because $C_p\times C_p$ has 'unexpected' normal subgroups.

Assume that $L>1$ and consider any $(a,b)\in L$, $(a,b)\neq 1$ Consider any $x\in H_2$. Then $(a,b)^{(1,x)} = (a,b^x)\in L$ because $L$ is normal. Therefore also $(a^{-1},b^{-1})(a,b^x) = (1,b^{-1}b^x)\in L$. But this implies that $b=b^x$.

And this for all $x\in H_2$, so $b$ would be central in $H_2$. This implies $H_2$ must be abelian, i.e. cyclic of prime order. Similarly $H_1$ is cyclic of prime order and then it's easy to finish with what you know about subgroups of $C_p\times C_p$ and $C_p\times C_q$.