# Inductive Step when working with Quotient Groups, Proof about normal subgroups of special groups

I have a question on the following proof of the characterisation of normal subgroups of groups which are direct products of non-abelian simple groups:

Theorem: Let $G = G_1 \times \cdots \times G_n$, where $G_1, \ldots, G_n$ are non-abelian and simple, and let $N \unlhd G$. Then there exists some set $J \subseteq \{1,\ldots, n\}$ such that $$N = \times_{j \in J} G_j \quad \mbox{ and } \quad G_k \cap N = 1 \mbox{ for } k \notin J.$$

The proof goes by induction on $n$. I just show the relevant part:

Suppose $k \in \{1,\ldots, n\}$ with $N\cap G_k \ne 1$. Then $G_k \le N$ because $G_k$ is simple., let $\overline G/G_k$, then $$\overline G = \times_{i\ne k} G_i/G_k$$ with $G_i / G_k \cong G_i$ for $i \ne k$. By induction hypothesis there exists $J' \subseteq \{1, \ldots, n \}$ with $k \notin J'$ such that $$\overline N = \times_{j\in J'} G_j/G_k \quad \mbox{ with } \quad G_i/G_k \cap \overline N = 1 \mbox{ for } i \notin j'.$$ But then for $J := J' \cup \{ k \}$ we have $$N = \times_{j\in J} G_j \quad \mbox{ with } \quad N\cap G_i = 1 \mbox{ for } i \notin J.$$

The last step I do not understand, the logic is something like ''because of $$G/N = G'/N$$ we can conclude $G = G'$'', but as I found out this logic does not hold in general, see the Extension problem. So whats going on here?

(remark on notation: do not know how to do the big $\times$...)

• As for notation, I think most people write \prod, which gives $G = \prod_{i=1}^n G_i$. – André 3000 Mar 12 '15 at 20:36
• Yes, but in the textbook they write it with the big $\times$, and $\prod_i G_i$ is simply $G_1G_2 \cdots G_n$, which need not be a direct product, just an ordinary product, so guess using your notation would be suboptimal. – StefanH Mar 12 '15 at 20:38
• It's probably overkill for this, but as far as your Extension problem remark goes: Ayoub proved that if $G=H\times K$ is a direct product of finite groups, then every short exact sequence $0\rightarrow H\rightarrow G\rightarrow K\rightarrow 0$ splits as a direct product. The proof can be generalized to groups with both chain conditions without a whole lot of difficulty, as an aside, and it has also been established for profinite groups by Guralnick et al., and for Chernikov groups by authors I can't remember. – zibadawa timmy Mar 12 '15 at 20:38
• If $N$ is a normal subgroup of a group $G$ and $N \le G_i \le G$ for $i=1,2$, then $G_1/N = G_2/N \Rightarrow G_1=G_2$. That follows immediately from the definition of quotient groups: $g \in G_1 \Rightarrow gN \in G_1/N \Rightarrow gN \Rightarrow G_2/N \Rightarrow g \in G_2$. – Derek Holt Mar 12 '15 at 20:40
• Before applying your inductive argument, you have to prove that there exists a $k$ with $N \cap G_k \ne 1$. – Derek Holt Mar 12 '15 at 20:43

You want to use the following result:

If $G$ is any group, $K$ is a direct factor of $G$, and $H$ is a subgroup of $G$ containing $K$, then $K$ is a direct factor of $H$.

In your case, you would use this to conclude that $G_k$ is a direct factor of $N$, and the inductive hypothesis tells you that the remaining factors (if any) are some (possibly empty) collection of the remaining $G_i$.

EDIT:

Proof of the claimed result:

Suppose $G=K\times L$ and $K\subseteq H\subseteq G$. For any $h\in H$ there exists $a\in K,b\in L$ with $h=ab$. Since $K\subseteq H$ we have $a^{-1}h = b\in H$. Setting $T=\{b\in L \ : \ ab\in H \mbox{ for some } a\in K\}$, we may conclude that $T$ is a subgroup of $H$ (and $L$, in fact) such that $H=KT$, $K\cap T=\{1_G\}$, and $[K,T]=\{1_G\}$. It follows that $H$ is the direct product of $K$ and $T$.

Note that you may use slightly different notation depending on how you opt to write out direct products like this. Namely you may write $h=(a,b)$ instead, in which case we'd then say $(a^{-1},1)h = (1,b)\in H$, etc. This is just a technical distinction between whether we interpret $G$, at the level of sets, as the actual cartesian product of the sets $K$ and $L$, or if we simply use the $\times$ to denote the direct product conditions without necessarily implying a Cartesian product of the underlying sets. The latter is the more general situation, and is what I was using.

In your case, we would thus have that $N=G_k\times L$ for some normal subgroup $L$ of $\times_{i\neq k} G_i$. Apply the inductive hypothesis to $L$. In this way we actually avoid the quotients entirely, thus avoiding your concerns about the extension problem. Though, see my comment about a result of Ayoub which shows the problem doesn't exist for direct products of certain groups.

• Thanks for your reply, but I still have several questions. First, strictly, by the isomorphism theorems we just could have $$\overline G \cong \times_{i\ne k} G_i/G_k$$ not equality! And if I got you right, this "inverse" application of the isomorphism of quotient groups just works if the groups involved are direct products? Seems to be a quite deep result, used in my introductory textbook, and being still in the first few chapters... also I guess all we can get is $N \cong \times_{j\in J} G_i$, so not equality, and this is considerable weaker as a group could contain many isomorphic copies? – StefanH Mar 13 '15 at 12:03
• @Stefan I've added proof details to my answer. In short, if the subgroup contains a direct factor then you can simply multiply out the terms from that factor and what's left is the commuting complement desired. – zibadawa timmy Mar 13 '15 at 20:46