conjugacy classes and order of group

Question

Suppost that $k_G(A)$ denotes the number of conjugacy classes of $G$ that intersects $A$ non-trivially ($A$ is an arbitrary subset of $G$) and $M=G^{'}Z(G)$. Also suppose that $G$ is non-solvable, $\frac{G}{M}$ is cyclic group of order 6 and $k_G(G-M)=5$. What can we say about $G$?

Answer 1

So all elements in each of the $5$ nontrivial cosets of $M$ in $G$ are conjugate. hence they all have centralizers of order $6$. So if $g \in G$ such that $gM$ generates $G/M$, then $g$ must have order $6$, and $G$ is a Frobenius group with kernel $M$ and complement $\langle g \rangle$.