# Minimal normal subgroups of Product of simple groups

Known Result:

Let

$$G= S_1 \times S_2 \times\dots\times S_n,$$

where each $$S_i$$ are non-abelian simple groups. Then $$S_i$$'s are the minimal normal subgroup of $$G$$. (Even $$S_i$$'s are the only minimal normal subgroups of of $$G$$.)

My claim is that this is true even when some $$S_i$$'s are abelian. More formally:

Consider

$$H= T_1 \times T_2 \times\dots \times T_m,$$

where each $$T_i$$'s are simple groups (abelian or non abelian).

Claim 1: Each $$T_i$$'s are minimal normal subgroups of $$H$$ (I mean the subgroups of $$H$$ isomorphic to $$T_i$$ is minimal normal in $$H$$).

Proof: Let $$K\leq H$$ such that $$K \cong T_i$$. $$K$$ is normal in $$H$$ by def of direct product.

If $$K$$ is not minimal normal in $$H$$, then there is $$N < K$$ such that $$N \trianglelefteq H$$. Thus there is $$T' \leq T_i$$ such that $$T' \cong N$$ and $$T' \trianglelefteq H$$. Which contradict the fact that $$T_i$$ is simple.

Thus $$K$$ is minimal normal in $$H.$$ And hence the claim follows.

EDIT: Question 1: Does the Claim 1 is correct?

Question2: If yes, are $$T_i$$'s only minimal normal subgroups of $$H$$.

• You will be aware of what happens when $G$ is the direct product of two (or more) groups of the same prime order. Commented Apr 1, 2022 at 13:14
• The statement for nonabelian $S_i$ is not merely that any minimal normal subgroup is isomorphic to o e of the $S_i$, but that it is exactly one of the $S_i$. Your "claim 1" incorrectly uses $\cong$, and the claim is false if you only assume $K$ is isomorphic to one of the $S_i$ Commented Apr 1, 2022 at 14:03
• @AndreasCaranti, I think Claim 1 is true in general. And the claim that 'these are the only minimal subgroups' of $H$ is not correct. As it may have diagonal subgroup which is minimal normal in $H$. E.g. $H= A_5 \times Z_3 \times Z_3$ has atleast four minimal normal subgroup which are $A_5, Z_3, Z_3$ and the diagonal subgroup $\langle (e, 1,1) \rangle$.
– Jins
Commented Apr 1, 2022 at 15:26
• The "diagonal subgroups" only occur when you have abelian factors. Commented Apr 1, 2022 at 15:33
• As written, Claim 1 is false. The subgroup of $A_5\times A_6$ that consists of the pairs $(x,y)$ where $y(6)=6$ and $y$ restricted to $\{1,2,3,4,5\}$ is the same as $x$ is isomorphic to $A_5$, but is not normal in $A_5\times A_6$. The problem is your use of "subgroups of $H$ isomorphic to $T_i$"; there may be other subgroups isomorphic to the factors, as noted already. And your claim that $K\cong T_i$ implies normality is false, as it is currently written. Commented Apr 1, 2022 at 21:47

You are being careless with equalities and isomorphisms.

Let $$G=S_1\times\cdots \times S_n$$ where each $$S_i$$ is simple. Let $$\iota_k\colon S_k\hookrightarrow G$$ be the canonical embeddings, sending $$g\in S_k$$ to the tuple $$(e,\ldots,g,\ldots, e)$$ that has $$g$$ in the $$k$$th coordinate.

Claim 1: For each $$k$$, $$\iota_k(S_k)$$ is a minimal normal subgroup of $$G$$.

This is true because each $$\iota_k(S_k)$$ is normal, and any normal subgroup of $$G$$ contained in $$\iota_k(S_k)$$ corresponds to a normal subgroup of $$\iota_k(S_k)\cong S_k$$, hence is trivial or all of $$S_k$$.

Claim 2: If each $$S_k$$ is nonabelian, then the only normal subgroups of $$G$$ are of the form $$\iota_{i_1}(S_{i_1})\cdots\iota_{i_r}(S_{i_r})$$ for some $$1\leq i_1\lt\cdots\lt i_r\leq n$$, $$0\leq r\leq n$$.

These are all normal, being products of normal subgroups. Now assume that $$N\triangleleft G$$. Then $$\pi_j(N)\triangleleft S_j$$, hence the projection is either trivial or all of $$S_j$$. Thus, we may assume that $$N$$ is a subdirect product $$G=S_1\times\cdots\times S_n$$, and we aim to show that it is then equal to $$G$$.

I claim that $$\iota_1(S_1)\subseteq N$$. Indeed, let $$x,y\in S_1$$. Because $$N$$ is a subdirect product, there exists an element in $$N$$ with first component equal to $$x$$, say $$(x,s_2,\ldots,s_n)\in N$$. Conjugating with $$\iota_1(y)=(y,e,e,\ldots,e)$$, we have that $$(yxy^{-1},s_2,\ldots,s_n)\in N$$. Now multiplying by $$(x,s_2,\ldots,s_n)^{-1}\in N$$ on the right we obtain $$(yxy^{-1}x^{-1},e,\ldots,e)= ([y,x],e,\ldots,e)\in N.$$ Thus, $$\iota_1([S_1,S_1])\subseteq N$$. But because $$S_1$$ is nonabelian, $$S_1=[S_1,S_1]$$; thus, $$\iota_1(S_1)\subseteq N$$.

Similarly, $$\iota_k(S_k)\subseteq N$$ for each $$k$$; thus, a generating set for $$G$$ is contained in $$N$$, so $$N=G$$, as desired.

Corollary. If all $$S_k$$ are nonabelian and simple, then the minimal normal subgroups of $$S_1\times\cdots \times S_n$$ are precisely the subgroups $$\iota_k(S_k)$$, for $$k=1,\ldots,n$$.

Claim 2 is false if you have two isomorphic abelian factors, as then you have "diagonal" subgroups, like $$\langle (x,x)\rangle\leq C_2\times C_2$$.

As stated your Claim 1 is false, because you may have subgroups isomorphic to some of the $$S_i$$ which are not normal. For example, if there is any $$S_k$$ that is isomorphic to (a subgroup of) $$S_j$$ for some $$k\neq j$$ via a morphism $$\phi$$, then the subgroup generated by the elements $$\iota_k(x)\iota_j(\phi(x))$$ is a subgroup that is isomorphic to $$S_k$$, but will not be normal in $$G$$.

Let's look a bit more carefully at what happens when you have abelian factors. Suppose that $$G=S_1\times\cdots\times S_n\times A_{n+1}\times\cdots\times A_{n+m}$$ where the $$S_i$$ are simple and nonabelian, and $$A_{n+1},\ldots,A_n$$ are simple abelian (hence cyclic of prime order).

Lemma. If $$N\triangleleft G$$ and there exists $$k$$, $$1\leq k\leq n$$ such that $$\pi_k(N)\neq\{e\}$$, then $$\iota_k(S_k)\leq N$$.

Proof. The argument is similar to the one for Claim 2 above: given $$x,y\in S_k$$, we have an element $$n\in N$$ with $$\pi_k(n)=x$$. Then $$\iota(y)n\iota(y)^{-1}n^{-1}$$ has $$k$$th entry equal to $$[y,x]$$, and all other entries trivial, so $$\iota_k([S_k,S_k])\leq N$$; but this equals $$\iota_k(S_k)$$ because $$S_k$$ is simple and nonabelian. $$\Box$$

Corollary. If $$N\triangleleft G$$, and $$(s_1,\ldots,s_n,a_1,\ldots,a_m)\in N$$, then $$(s_1,\ldots,s_n,e,\ldots,e)\in N$$ and $$(e,\ldots,e,a_1,\ldots,a_m)\in N$$.

Proof. If any $$s_j\neq e$$, then $$N$$ contains $$\iota_j(S_j)$$, Thus, it contains $$(s_1,\ldots,s_n,e,\ldots,e)$$. $$\Box$$

Corollary. If $$N\triangleleft G$$, then $$N=M\times B$$, where $$M\triangleleft S_1\times\cdots\times S_n$$, and $$B\triangleleft A_{n+1}\times\cdots\times A_{n+m}$$.

Now, we already know what the normal subgroups of $$S_1\times \cdots\times S_n$$ are, so we know what $$M$$ looks like.

For $$B$$, we have a finite abelian group, which is isomorphic to $$E_{p_1,k_1}\times\cdots\times E_{p_r,k_r}$$, where $$p_1,\ldots,p_r$$ are pairwise distinct primes, $$k_i\geq 1$$, and $$E_{p_i,k_i}$$ is an elementary abelian $$p_i$$-group of rank $$k_i$$; that is, $$E_{p_i,k_i}\cong (C_{p_i})^{k_i}$$. We know that $$B$$ is the product of its $$p_i$$ parts, so we are left with considering the subgroups of $$E_{p_i,k_i}$$. From linear algebra we know they just elementary abelian, isomorphic to $$(C_{p_i})^{t_i}$$ with $$0\leq t_i\leq k_i$$.

In particular, any minimal normal subgroup will be isomorphic to $$C_{p_i}$$, hence to one of the factors.

So we have:

Theorem. Let $$G= S_1\times\cdots \times S_n$$, where each $$S_i$$ is a finite simple group. Then the minimal normal subgroups of $$G$$ are precisely the subgroups $$\iota_k(S_k)$$ if and only if for each prime $$p$$ there is at most one $$S_i$$ that is cyclic of order $$p$$. If there are other minimal normal subgroups of $$G$$, then they are isomorphic to a cyclic group of order $$p$$, and so are at least two of the $$S_k$$.

• Thanks for the clarification. Can you clear the following doubt. In $H$ at least one $T_i$ is non abelian, others can be any simple group. In this case the diagonal subgroup will not be minimal normal and hence the only minimal normal subgroups are precisely $T_i$'s.
– Jins
Commented Apr 1, 2022 at 16:53
• @Jins: First: you need to be absolutely clear about whether you mean the images of the $T_i$ under the canonical embeddings into theproduct, or just some subgroup isomorphic to the $T_j$. You already have examples of normal subgroups that are not just the images of the direct factors under the canonical embeddings when you have at least two abelian factors, even if at least one of the $T_i$ is nonabelian. Commented Apr 1, 2022 at 17:51