# Finite groups with unique subgroup of order $p$, for all prime divisors of $G$

Let $G$ be a finite group (non $p$-group) with the following properties:

a: For every prime divisor $p$ of $\vert G\vert$, there exists only one minimal subgroup of order $p$.

b: For every pair of distinct prime divisors of $\vert G\vert$ say $p$ and $q$, If $P_{1}, P_{2},..., P_{m}$ are all of the $p$-subgroups of $G$ and similarly $Q_{1}, Q_{2},..., Q_{n}$ are all of the $q$-subgroups of $G$, then $G$ has $mn$ subgroups of order $p^{\alpha}q^{\beta}$($\alpha,\beta\geq 1$). ( If $\vert G\vert=p^{s}q^{t}$ for some naturals $s,t$, then we count it in this computation )

Can we say that $G$ is a Dedekind group.( i.e. is every subgroup of $G$, a normal subgroup? )

• Are those different questions or are you asking for group that satisfy both a. and b.? – Johannes Hahn Sep 23 '16 at 21:45
• no I edited it, because I found a counter example for previous version of my question. So I changed a bit in the conditions. – H.Shahsavari Sep 23 '16 at 22:47

## 2 Answers

I believe that it is possible to prove by elementary methods that a finite group $G$ which has a unique subgroup of order $p$ for each prime divisor $p$ of its order is solvable. Since the derived $[G,G]$ inherits the property, (though its order may have fewer prime divisors), we may suppose that $G= [G,G]$, in which case every subgroup of prime order of $G$ is central. By elementary transfer, $G$ has a normal $p$-complement whenever $p$ is an odd prime. Then $G$ has a normal Sylow $2$-subgroup, and $G/O_{2}(G)$ is nilpotent, so $G$ is solvable. Once we know that $G$ is solvable, it is easy to obtain further restrictions on the structure of $G$, as $F(G)$ has transparent structure.

No. For example, the generalized quaternion groups $Q_{2^{n}}$ have only one subgroup of order $2$, but they are not Dedekind groups. For this reason, the group $Q_{16} \times \mathbb{Z}/3\mathbb{Z}$ is a group of order $48$ that is not a Dedekind group, but satisfies your conditions (a) and (b).

Condition (a) alone implies that every Sylow $p$-subgroup is either cyclic or generalized quaternion (because these are the only $p$-groups that have a unique subgroup of order $p$). Apparently, Zassenhaus (in the 1930s) classified all solvable groups where all the Sylow $p$-subgroups for odd primes are cyclic, and where the Sylow 2-subgroup contains an index $2$ cyclic subgroup (and Michio Suzuki, in the paper ''On finite groups with cyclic Sylow subgroups for all odd primes'' from Amer. J. of Math in 1955 classified the non-solvable groups with the same properties).

This gives a complete classification of the groups that meet your condition (a).

• Dear Jeremy. How could you prove that there does not exist more than one $p$-sylow subgrop for every prime dividing $G$? what can we say, If we replace the result with this claim that " Every finite group with these condition is a Dedekind group or direct product of a generalized quaternion group and a cyclic group of some odd order. Can we still give a counter example? Thank you for your concern – H.Shahsavari Sep 24 '16 at 0:57