I am trying to understand the structure of normal groups of $G=A\times H$ where $A$ is an abelian group and $H$ is a direct product of non-abelian simple groups (both are finite).

I want to show that if $N \vartriangleleft G$ then $N = N_A \times N_H$ where $N_A \vartriangleleft A$ and $N_H \vartriangleleft H$. But I don't have any proof or counterexample for this argument.

Is this argument true? How can I prove it?

  • 3
    $\begingroup$ $C_2 \times S_3$ has a normal subgroup of index $2$ that is not of that form. I don't think your assumptions on $A$ and $H$ imply anything in particular about the normal subgroups. $\endgroup$ – Derek Holt Apr 24 at 7:41
  • $\begingroup$ I've change the properties of $H$ to be a direct product of simple non-abelian groups. I hope this would give me what I need $\endgroup$ – roy999 Apr 24 at 9:32
  • 2
    $\begingroup$ Yes it's true under the new hypotheses. In fact if $G = A \times H_1 \times \cdots \times H_k$ with each $H_i$ nonabelian simple, then any normal subgroups is the direct product of its intersections with the factors (no need to assume that $G$ is finite). Perhaps try proving it first when $A$ is trivial. $\endgroup$ – Derek Holt Apr 24 at 9:49

Suppose that $G=S_1\times\cdots\times S_k$ is a direct product of nonabelian simple groups and let $N\triangleleft G$. I claim that $N$ is the direct product of some of the $S_i$; that is, $N=N_1\times\cdots\times N_k$, where $N_i$ is either trivial or equal to $S_i$ for each $i$. Note that if $\pi_j\colon G\to S_j$ is the projection on the $j$th coordinate, then $\pi_j(N)$ is normal in $S_j$, and hence $\pi_j(N)=\{e\}$ or $\pi_j(N)=S_j$ for each $j$. Moreover, we always have that $$N\subseteq \pi_1(N)\times\cdots\times\pi_k(N).$$

To verify the initial claim about $N$, first let $(g_1,\ldots,g_n)\in N$ be nontrivial; say that $g_1\neq e$. Since $S_1$ is simple and nonabelian, $g_1$ is noncentral, so there exists $x\in S_1$ such that $xg_1x^{-1}\neq g-1$. Now, we have that $$ (xg_1x^{-1},g_2,\ldots,g_n) = (x,e,\ldots,e)(g_1,g_2,\ldots,g_n)(x,e,\ldots,e)^{-1} \in N$$ and hence $$(xg_1x^{-1}g_1^{-1},e,\ldots,e) = (xg_1x^{-1},g_2,\ldots,g_n)(g_1,g_2,\ldots,g_n)^{-1}\in N.$$ Note that $xg_1x^{-1}g_1^{-1}\neq e$. Thus, $N\cap S_1\times\{e\}\times\cdots\times\{e\}$ is nontrivial. But this intersection is of the form $H_1\times\{e\}\times\cdots\times\{e\}$, with $H_1\leq S_1$. And since $N\triangleleft G$, it follows that $H_1\triangleleft S_1$. The simplicitly of $S_1$ now yields that $H_1=S_1$. Thus, $S_1\times\{e\}\times\cdots\times\{e\}\subseteq N$.

Repeating this argument for each $i$ such that $\pi_i(N)$ is nontrivial (where $\pi_i\colon G\to S_i$ is the projection onto the $i$th component) yields that if $\pi_i(N)$ is nontrivial, then $N$ contains $S_i$.

Since $N\subseteq \pi_1(N)\times\cdots\times\pi_k(N)$, this yields the conclusion that $N$ is of the desired form.

Now suppose that $G=A\times S_1\times\cdots\times S_k$, with $A$ abelian and $S_i$ nonabelian and simple for all $i$. Let $N\triangleleft G$.

Proceeding as above, with $\pi_i\colon G\to S_i$ the projection onto $S_i$ we conclude that if $\pi_i(N)\neq\{e\}$, then $N$ contains all elements with $j$-coordinate trivial for $j\neq i$.

Now consider an arbitrary element $(a,s_1,\ldots,s_k)\in N$. Then $(e,s_1,\ldots,s_k)\in N$, and hence $(a,e,\ldots,e)\in N$. Thus $N$ also contains all elements with trivial $S_i$ coordinate, and any element of $\pi_A(N)$ in the $A$-coordinate. This shows that we do indeed have $$ N = \pi_A(N)\times\pi_1(N)\times\cdots\times \pi_k(N),$$ where $\pi_i(N)$ is either trivial or equal to $S_i$ for each $i$.


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