# Normal group of product of abelian and non-abelian simple groups

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?

• $C_2 \times S_3$ 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. – Derek Holt Apr 24 at 7:41
• 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 – roy999 Apr 24 at 9:32
• 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. – 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 $$i$$.