Assume $H,G$ are simple groups. Can we prove that either $H\unlhd HG$ or $H\unlhd GH$?

Question

Assume $H,G$ are simple groups. Can we prove that $H$ is normal in $HG$ or in $GH$?

Context:

I want to use this argument in order to prove the following statement:

If $H_i$ and $G_j$ are simple such that $G_1 \times...\times G_n $ isomorphic to $H_1\times... \times H_m $ Then $n=m$ and by reindexing we can get that $G_i$ is isomorphic to $H_i$ for any $i$.

I want to claim that

$$H_1 \trianglelefteq H_1H_2 \trianglelefteq ... \trianglelefteq H_1H_2...H_m $$

Is a composition series of $H_1...H_m$, and then I'll use Jordan-Holder theorem (somehow, did not figured out the details yet) to conclude what I wanted to prove.

Thanks in advance.

Answer 1

No. Any transposition of $S_3$ generates a simple subgroup that is not normal in $S_3$, even though $S_3$ is the product of the subgroup generated by any transposition and the subgroup generated by any $3$-cycle.

Answer 2

If $H_1$ and $H_2$ are simple subgroups of a larger group $G$, and $H_1 H_2$ is a subgroup of $G$ it is not true in general that $H_1$ and $H_2$ are both normal in $H_1H_2$. An example might be

$$G = \{ \begin{pmatrix} x & y \\ & 1 \end{pmatrix} : 0 \neq x \in \mathbb F_3, y \in \mathbb F_3\}$$

$$H_1 = \{ \begin{pmatrix} x \\ & 1 \end{pmatrix} 0 \neq x \in \mathbb F_3\}$$

$$H_2 = \{ \begin{pmatrix} 1 & y \\ & 1 \end{pmatrix} y \in \mathbb F_3\}$$

Here $\mathbb F_3$ is the field with three elements, $H_1 \cong \mathbb Z/2\mathbb Z$ and $H_2 \cong \mathbb Z/3\mathbb Z$ are both simple groups, and $H_1H_2 = G$ is a group, yet $H_1$ is not normal in $H_1H_2$.

But in your case, of course $H_1$ is normal in $H_1 \times H_2$, $H_1 \times H_2$ is normal in $H_1 \times H_2 \times H_3$, and so on. This gives as you say a composition series whose quotients are the simple groups $H_1,H_2, ...$. You can use Jordan-Holder to argue that $n=m$ and that with some reordering, each $H_i \cong G_i$.

$$\begin{pmatrix} 1 & y \\ & 1 \end{pmatrix} \begin{pmatrix} x \\ & 1 \end{pmatrix} \begin{pmatrix} 1 & -y \\ & 1 \end{pmatrix}$$

$$\begin{pmatrix} x & y \\ & 1 \end{pmatrix}\begin{pmatrix} 1 & -y \\ & 1 \end{pmatrix} = \begin{pmatrix} x & y-xy \\ & 1 \end{pmatrix}$$