-3
$\begingroup$

if $H$ and $K$ are nonabelian simple groups prove that :

$H$ $\times$ $K$ has exactly four distinct normal subgroups.

Please help me prove this.

  • 1
    Do you have any thoughts about what those four subgroups might be? – 2017-02-22

2 Answers 2

1

I assume for simplicity that this is an internal direct product, so that $H$ and $K$ are normal subgroups of the group $G = H \times K$.

I claim that either $L \cap H \ne 1$ or $L \cap K \ne 1$. In fact if $1 \ne (h, k) \in L$, with $h \ne 1$, since $H$ is non-abelian, there is $t \in H$ such that $h^{t} \ne h$. Then $(h, k)^{(t, 1)} = (h^{t}, k) \in L$, and $(h, k)^{-1} \cdot (h^{t}, k) = (h^{-1} h^{t}, 1)$ is a non-trivial element of $L \cap H$. A similar argument applies if $k \ne 1$.

Suppose then for instance that $L \cap H \ne 1$. Since $1 \ne L \cap H \trianglelefteq H$, and $H$ is simple, we have that $L \cap H = H$, so $L \ge H$, and since $L/H$ is a normal subgroup of the simple group $G/H \cong K$, then either $L = H$ or $L = G$.

  • 0
    thanks for your answer! where are subgroups ? H and K and another ??! why subgroups are distinct ? – 2017-02-22
  • 0
    i want to find 4 normal subgorups of G where are distinct... please help – 2017-02-23
0

Hint: put $M=H \times \{1\}$, and $N=\{1\} \times K$. Then $M,N \unlhd G$, $G=MN$, and $M \cap N=\{(1,1)\}$. Now if $T \unlhd G$, what can you say about $T \cap M$ and $T \cap N$?

  • 0
    thanks for your answer ! $T$ $\cap$ $M$ and $T$ $\cap$ $N$ are normal subgroups of G ! but how to prove there are " exactly " four subgorups ? – 2017-02-22
  • 0
    Yes they are even normal subgroups of $M$ and $N$ respectively. But these groups are simple, so ... – 2017-02-22
  • 0
    why this subgroups are distinct? – 2017-02-22