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

Question

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.

Answer 1

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 $1 <H_1 < H_2 < \cdots < H_k=G$ 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.