Let $\mathcal{FG}$ be the category of finite groups. Let $S$ be a full subcategory of $\mathcal{FG}$.
Assume that $G\in \mathcal{FG}$ and $P\in S$ is a subgroup of $G$. We say that $P$ is $S$-maximal if there is no object $P'\in S$ with $P\subset P' \subset G$. Assume that $S$ satisfies the following conditions:


1. The subcategory $S$ is closed under taking subgroups, taking extensions" and isomorphy. That is: $P\in S, Q\subset P$ implies that $Q \in S$. Moreover for every short exact sequence $1\to P\to Q\to R\to 1$ we have $Q\in S$ if $P,R \in S$. Moreover every group isomorphic to an object of $S$ lies in $S$.
2. For every $G\in \mathcal{FG}$, every two $S$-maximal subgroups of $G$ are conjugate. Moreover for every maximal $S$-subgroup $P$ of $G\in \mathcal{FG}$ we have $N(N(P))=N(P)$ and $\lvert G\rvert/\lvert N(P)\rvert$ is coprime to $|P|$.


Does it imply that $S$ is the category of $p$-groups for some prime number $p$?



A possible characterization of the category of finite $p$-groups
We can get most of the way to an answer, without using all the conditions.


**Definition:** Say that $S \subseteq \mathcal{FG}$ is *nice* if it is nonempty and closed under subgroups, extensions, and isomorphy, and has the property that for any $G \in \mathcal{FG}$, the $S$-maximal subgroups of $G$ are all conjugate.


**Proposition:** Suppose that $S \subseteq \mathcal{FG}$ is nice. Then there is a set $\mathcal P$ of primes such that $P \in S$ if and only if the prime divisors of $|P|$ are all in $\mathcal P$.


This implies that any [$\mathcal P$-Hall subgroup](https://en.wikipedia.org/wiki/Hall_subgroup) of $G$ is a maximal $S$-subgroup of $G$. Conversely, if $G$ has a maximal $S$-subgroup $H$, then assuming niceness, since $H$ contains the Sylow $p$-subgroups for each $p \in \mathcal P$, we have that $H$ is a $\mathcal P$-Hall subgroup. That is, if $S$ is nice, then a maximal $S$-subgroup is precisely a Hall $\mathcal P$-subgroup.


**Proof:** Let $\mathcal P$ be the set of primes $p$ such that there exists $P \in S$ with $p$ dividing $|P|$ (as it must be). By passing to subgroups, we have $C\_p \in S$ for $p \in \mathcal P$, and by taking extensions we have that every finite $p$-group is in $S$ for every $p \in \mathcal P$. Let $G$ be a finite group such that the prime divisors of $|G|$ are in $\mathcal P$. Now, there are $S$-maximal subgroups of $G$ containing each Sylow subgroup of $G$, and by hypothesis these are all conjugate and in particular have the same order. So if $|G| = p^n m$ with $p$ not dividing $m$, then $p^n$ divides the order of any $S$-maximal subgroup $P \subseteq G$. Since this is true for all prime divisors $p$ of $|G|$, we have that $|G|$ divides $|P|$, so that $P = G$, and thus $G \in S$.




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It's known that not all $\mathcal P$-Hall subgroups of a finite group $G$ are conjugate in general, but I don't happen to know whether the known counterexamples cover all possible sets $\mathcal P$ of primes other than the subsingletons and the set of all primes. If $G$ is solvable, then its $\mathcal P$-Hall subgroups *are* conjugate -- so if we replace $\mathcal{FG}$ with just the solvable finite groups we get an expanded set of possibilities for $S$.




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(Many of the comments relate to an earlier version of this post which attempted to whittle down the set of $\mathcal P$ allowed. Thanks to Richard Lyons below for pointing out the mistake.)




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**EDIT:** I just wanted to add another way to close the argument, alternate to Richard Lyons' above, which relies on the original literature on Hall subgroups.


Hall showed in [Theorems Like Sylow's](https://doi.org/10.1112/plms/s3-6.2.286), discussion after Theorem A4, that if $\mathcal P$ contains two primes smaller than $n$, then any Hall $\mathcal P$-subgroup of $\Sigma\_n$ cannot be solvable. Then Thompson showed in [Hall subgroups of the symmetric groups](https://doi.org/10.1016/S0021-9800(66)80032-7) that the only nonsolvable Hall subgroups of $\Sigma\_n$ are the trivial group, $\Sigma\_n$ itself, and (if $n$ is prime) $\Sigma\_{n-1}$. So if $\mathcal P$ contains $p,q$ and omits $r$, then take $G = \Sigma\_n$ for any composite $n > p,q,r$. The only candidate Hall subgroups of $G$ are the trivial group and $G$ itself; the former is not a $\mathcal P$-Hall subgroup because $p | n!$ and the latter is not because $r | n!$.


