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.



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


