# Assigning a finite number, $n(G)$, to a finite group $G$ with this property

I want to assign a finite number, $$n(G)$$, to a finite group $$G$$ such that if $$H$$ is a proper retract of $$G$$, then $$n(H)\lneq n(G)$$. By a retract of $$G$$, I mean a subgroup $$H$$ of $$G$$ for which there is an epimorphism $$r:G\to H$$ such that $$r(x)=x$$ for all $$x\in H$$. By a proper retract, I mean a retract $$H$$ such that $$H\neq G$$.

I want that this number depends on the group properties of $$G$$. So the cardinality of $$G$$ is not a good idea for me. I think $$n(G)$$ will be in such a way that if $$G$$ is a finite abelian group, then it coincides with the number of non-zero direct summands of $$G$$. (In fact, if $$G$$ is a finite abelian group and $$H$$ is a proper retract of it (which is a direct summand and vice versa), then by the fundamental theorem of finitely generated abelian groups we get that the number of direct summands of $$H$$ is less than the number of direct summands of $$G$$.)

My idea: If $$H$$ is a retract of $$G$$, then $$G=H\ltimes N$$, where $$N$$ is a normal subgroup of $$G$$. For example, $$S_2 =\mathbb{Z}_2\ltimes \mathbb{Z}_3$$. I can say that $$n(S_3)=2$$; I mean the number of semidirect summands (both normal and non-normal ones). But I don't how to define it for bigger groups.

• The length of a chief series of $G$ would work, but that is still too crude for abelian groups. Perhaps something like the maximum length of a normal series of $G$ in which all subgroups are complemented? Commented May 22, 2023 at 15:56
• @DerekHolt That is a good idea. Could you please guide me how to prove that this number satisfies $n(H)\lneq n(G)$? Commented May 22, 2023 at 16:04
• @DerekHolt Is "normal" necessary? Retracts are not necessarily normal. Their complement are normal. Commented May 23, 2023 at 13:13
• If $H$ is a proper retract of $G$, then $H$ has a normal complement $N$ in $G$. So there is a complemented normal series of $G$ that includes $N$ together with the subgroups $NH_i$, where $H_i$ is a complemented normal series of $H$. So we get $n(G) \ge n(H)+1$. Commented May 23, 2023 at 13:45
• @DerekHolt Thank you very much for your help. I understood you comment in this way: ‎Assume that ‎‎$‎‎H=H_0 >H_1 >‎\cdots ‎>H_{n(H)}=1‎$ is a complemented normal series of ‎$‎‎H$ ‎with ‎the‎ ‎maximum ‎length ‎$‎‎n(H)$.‎ Since ‎‎$‎‎H$ ‎is a‎ ‎proper ‎retract ‎of ‎‎$‎‎G$, ‎‎it has a normal complement ‎$‎‎N$‎ in ‎$‎‎G$. ‎Then ‎ $G=G_0=N‎‎H=NH_0 >NH_1 >‎\cdots ‎>NH_{n(H)}=N>1‎$ is a complemented normal series of ‎$‎‎G$. and so ‎$n(G)\geq n(H)+1‎‎$.‎ Commented May 25, 2023 at 13:02

As you said yourself, you could define $$n(G) = |G|$$, but you are not happy with that. I am guessing that you would like $$n(G)$$ to be as small as possible. The length of a chief series of $$G$$ would also work, and would be an improvement on $$|G|$$, but we can do better than that.

I propose to define $$n(G)$$ to be the maximum length of a strict splitting normal series of $$G$$.

To be more precise, a normal series for $$G$$ is a series $$1 \le N_1 \le N_2 \le \cdots \le N_k = G$$ in which each $$N_i$$ is a normal subgroup of $$G$$, it is strict if all of the inclusions are strict, and splitting if each $$N_i$$ has a complement in $$G$$.

If $$H$$ is a proper retract of $$G$$ then it has a nontrivial normal complement $$N$$ in $$G$$ and then (as discussed in the comments) if $$1 < H_1 < \cdots < H_{n(H)} = H$$ is a maximum length strict splitting normal series of $$H$$, then $$1 < N < NH_1 < \cdots< NH_{n(H)} = G$$ is a splitting normal series for $$G$$, so we get $$n(G) \ge n(H)+1$$, and the function $$n(G)$$ has the required property.

Note that this function gives the optimal answer for abelian groups of the number of direct factors in a conical decomposition.

You proposed an alternative definition $$n'(G)$$ as the maximum length of a strict series $$1 of subgroups in which each $$H_i$$ with $$i < k$$ is a proper retract of $$G$$. That would also work, and I claim that $$n'(G)=n(G)$$.

It is easy to see that a strict splitting normal series gives rise to a strict series of proper retracts, so we get $$n(G) \le n'(G)$$.

It is less clear that a strict series $$1 =H_k < H_{k-1} < \cdots < H_1 < H_0=G$$ of proper retracts gives rise to a strict splitting normal series, but we can see that as follows.

Let $$N_i$$ be a normal complement of $$H_i$$ in $$G$$. We do not necessarily have $$N_i < N_{i+1}$$, but we can redefine the $$N_i$$ to get that property. Since $$H_2 < H_1$$ and $$G=H_2N_2$$, we get $$H_1 = H_2(N_2 \cap H_1)$$. Then $$N_1(N_2 \cap H_1)$$ is also a normal complement of $$H_2$$ in $$G$$, so we can redefine $$N_2 = N_1(N_2 \cap H_1)$$, etc.

Hence $$n'(G) \ge n(G)$$ and we have $$n(G)=n'(G)$$ as claimed.

• Thank you so much for your nice answer and help. I've learned good points from your comments and answer. Appreciate it. Commented May 27, 2023 at 3:32
• Dear @DerekHolt , is the proof of this statement "It is easy to see that a strict splitting normal series gives rise to a strict series of proper retracts" similar to what you proved about "a strict series of proper retracts gives rise to a strict splitting normal series"? Commented May 27, 2023 at 3:35
• I am really sorry to ask so many questions. Just my last question is: Why do you prefer the complemented normal series version? Commented May 27, 2023 at 3:37
• @DerekHolt Is $N_1 (N_2 \cap H_1)$ normal in $G$? Commented May 27, 2023 at 10:19
• Yes, it is normal in $G$. I guess I prefer the splitting chief series definition because it does not explicitly mention retracts, but since it is equivalen to your definition it doesn't matter much. Commented May 27, 2023 at 11:16