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? )



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


