# Finite nonabelian groups with abelian maximal subgroups

Please improve the title if it is not suitable.

Does there exist a finite group $G$ with the following properties?

$G$ has a minimal subgroup say $L$ isomorphic to $\mathbb{Z}_{p}$ and a maximal subgroup say $M$ isomorphic to $\mathbb{Z}_{pq}$ containing properly $L$ ($p$ and $q$ are some distinct primes), such that $L\ntrianglelefteq G$, $M\ntrianglelefteq G$ and $M$ is the only proper subgroup of $G$ containing properly $L$.

• It is clear that every Sylow $p$-subgroup of $G$ is of order $p$. Otherwise the subgroup $L$ is contained properly in a Sylow $p$-subgroup of $G$ which is obviously not equal to $M$, a contradiction. So $N_{G}(L)=M$ and by by Burnside's normal $p$-complement theorem we see that there exists a proper normal subgroup of $G$ say $N$ such that $(\vert N\vert,p)=1$. We claim that $N$ is a $q$-group. Obviously $q\mid\vert N\vert$. Contrary suppose that $r$ is a prime number other than $q$ such that $r\mid\vert N\vert$ and let $R$ be a Sylow $r$-subgroup of $G$. If $R\trianglelefteq G$, then $LR$ ... – H.Shahsavari Aug 9 '17 at 4:27
• then $LR$ is a proper subgroup $G$ containing properly $L$, a contradiction. So $R\ntrianglelefteq G$. Since $R\leqslant N\trianglelefteq G$, then all Sylow $r$-subgroups of $G$ are contained in $N$. Therefore number of Sylow $r$-subgroups of $G$ is equal to those that are Sylow $r$-subgroups of $N$ too. Thus we have $[G:N_{G}(R)]=[N:N_{N}(R)]$. But this is impossible since $p$ divides the left side of the equation while obviously it does not divide the right side. – H.Shahsavari Aug 9 '17 at 4:41
How about $G = {\rm SL}(2,3) = Q_8.C_3$ with $|L|=3$ and $|M|=6$?
There are similar examples of the form $N.C_p$, with $N$ an extraspecial $q$-group exponent $q$, where the $C_p$ subgroup acts irreducibly on $N/Z(N)$.