Let $G$ be a finite group and $M$ be a nontrivial proper subgroup of $G$ with the following conditions:


a) If $H$ is a subgroup of $G$ such that $M\lneqq H\lneqq G$, then $H$ contains at least one minimal subgroup of $G$ say $L$, such that $M\cap L=1$.


b) If $K$ is a subgroup of $G$ such that $M\cap K=1$ then $K\cong Z\_p$ for some prime $p$.


Can we say that $M$ is a maximal subgroup of $G$?


Please read my comments after Derek Holt's answer.



Maximal Subgroups
I remember answering a very similar (possibly the same) question recently but I cannot find it. 


Let $G$ be a central product of a cyclic group $M$ of order $4$ and the dihedral group $D\_8$ of order $8$. So $|G|=16$. (In fact you get an isomorphic group if you replace $D\_8$ by $Q\_8$. A central product of $C\_4$ with an extraspecial $2$-group is called a group of *symplectic type*.)


Then all subgroups of $G$ strictly between $M$ and $G$ have order $8$ and are isomorphic to $C\_4 \times C\_2$, and have the $C\_2$ as the required subgroup $L$. But $M$ has no complement in $G$, so any subgroup $K$ with $K \cap M=1$ has order $1$ or $2$. (You need to assume that $K$ is nontrivial in the problem.) This example also satisfies your condition c), since the subgroups $D\_8$ (and $Q\_8$) are maximal and contain all minimal subgroups of $M$.


You can construct similar examples for any prime $p$ as a central product of $C\_{p^2}$ with an extraspecial group of order $p^3$.


