# Finite groups with all maximal subgroups isomorphic

Suppose $$G$$ is a finite group such that any two maximal subgroups of $$G$$ are isomorphic. What can be said about such a group? Can they be classified?

The finite groups that have a unique maximal subgroup are exactly the cyclic groups of prime power order, $$\mathbb{Z}/p^n\mathbb{Z}$$ - these are the simplest examples of such groups. Also powers of $$\mathbb{Z}/p^n\mathbb{Z}$$, i.e. $$(\mathbb{Z}/p^n\mathbb{Z})^m = \mathbb{Z}/p^n\mathbb{Z} \times \ldots \times \mathbb{Z}/p^n\mathbb{Z}$$, have this property, and I think this covers all abelian groups with this property.

In general I think such a group has to be a $$p$$-group, by considering maximal subgroups containing Sylow subgroups for different primes.

This paper https://bib.irb.hr/datoteka/402744.SiCh.pdf calls such groups isomaximal, but seems to only handle $$2$$-groups up to order $$64$$.

Further question: what about groups $$G$$ such that any two maximal subgroups are isomorphic under some automorphism of $$G$$ (i.e. $$\operatorname{Aut}(G)$$ acts transitively on the set of maximal subgroups)? (Note: if this is strengthened to any two maximal subgroups being conjugate, then by this answer it becomes the same as having a unique maximal subgroup.)

• I agree with your sketch that such a group must be a p-group. – Josh B. Feb 6 '20 at 22:55

No idea about a classification, but here are some non-abelian examples. First recall that in a finite $$p$$-group, the maximal proper subgroups are just the kernels of homomorphisms onto $$\mathbf{Z}/p\mathbf{Z}$$.
(1) the free group of rank $$k$$ in the variety $$V_{p,k,\ell}$$ of $$k$$-step nilpotent groups satisfying the law $$x^{p^\ell}=1$$. In this case the automorphism group acts transitively on the set of maximal proper subgroups.
(2) the Heisenberg group of order $$p^{2n+1}$$ (at least for odd $$p$$). It can be described as the set of square matrices of size $$n+2$$ of the form $$\begin{pmatrix} I_1 & x & z \\ 0 & I_n & y \\ 0 & 0 & I_1\end{pmatrix}$$ over $$K=\mathbf{Z}/p\mathbf{Z}$$. In this case also, the automorphism group acts transitively on the set of maximal proper subgroups. I haven't checked $$p=2$$, and also I haven't checked if it works with fields $$\mathbf{F}_{p^k}$$ for $$k\ge 2$$.
There are many variants of the examples in (1) (using other varieties, e.g., exponent $$p^2$$ with derived subgroup of exponent $$p$$ etc.) Among groups in (2) there are groups of exponent $$p$$ and order $$p^3$$ and $$p^5$$, and among those in (1) there are groups of exponent $$p$$ and order $$p^5$$ and $$p^6$$.
I don't know if there are non-abelian groups of order $$p^4$$ answering the question but this should be checkable. Also I don't know examples for which the automorphism group doesn't act transitively on maximal subgroups. (Edit: there are such examples for both, see the comments.)
• Regarding groups of order $p^4$: at least for $p = 2$, the nontrivial semidirect product $\mathbb{Z}/4\mathbb{Z} \rtimes \mathbb{Z}/4\mathbb{Z}$ is an example (this is mentioned in the paper cited above). This is also an example where the stronger condition (automorphism group acts transitively) fails, since one of the maximal subgroups is characteristic. – math54321 Feb 7 '20 at 1:11
• @math54321 indeed, more generally $\mathbf{Z}/p^2\mathbf{Z}\rtimes \mathbf{Z}/p^2\mathbf{Z}$ (where the generator acts by multiplication by $1+p$) works for every prime $p$: every subgroup of index $p$ is abelian, isomorphic to $\mathbf{Z}/p^2\mathbf{Z}\times \mathbf{Z}/p\mathbf{Z}$, and moreover has 2 orbits of such subgroups modulo automorphisms, one being reduced to the characteristic subgroup containing the normal copy of $\mathbf{Z}/p^2\mathbf{Z}$. – YCor Feb 7 '20 at 9:31