A finite group $G$ satisfies property $P\_n$ if for every prime integer $p$, $G$ has at most $(n−1)$ non-central conjugacy classes the order of the representative element of which is a multiple of $p$. Let for $x\in G$, $x^G$ be a conjugacy classe of $G$. Also Let $G$ be a solvable group satisfying property $P\_5$, $M=G^{'}Z(G)$, $|\frac{G}{M}|=2$ and $G-M=x^G\cup y^G\cup z^G$, where ‎$|C\_G(x)|=|C\_G(y)|=|C\_G(z)|=6$‎, ‎$o(x)=2$‎, ‎$o(y)=6$ and $z=y^{-1}$‎.‎ Also ‎$‎M‎$‎ is of odd order and ‎$‎M‎$‎ has a normal and abelian 3-complement ($G^{'}$ is the commutator subgroup and $Z(G)$ is the center of $G$). Can we say that $Z(G)=1$?



Finite groups $G$ satisfying property $P_n$
There are no finite groups $G$ such that all your assumptions are met (so in a sense, the answer is "Yes"). 


Let $A$ be the normal $3$-complement of $G$, so that $M/A$ is a $3$-group. I claim that $|M/A| = 3$. Otherwise, the number of conjugacy classes of $M/A$ is at least $9$. Since $|C\_G(x)|=6$, we have $|C\_M(x)|=3$ and thus $M/A$ has at least $6$ conjugacy classes which lift to conjugacy classes not contained in $Z(G)$. Since $|G/M|=2$, the subgroup $M$ contains at least $3$ non-central conjugacy classes of $G$, such that its elements have order divisible by $3$. Since also $y^G$ and $z^G$ are such conjugacy classes, this contradicts property $P\_5$. 


Thus $|M/A| = 3$, and $G = \langle y \rangle A$. Therefore, $G' \subseteq A$ and we must have $y^2 \in Z(G)$ (because otherwise $M\subseteq A$.) But this also leads to a contradiction: For each $a\in A\setminus 1$ we get two further non-central conjugacy classes of order-$3$-elements, namely $(y^2a)^G = \{y^2a, y^2a^x\}$ and $(y^4a)^G=\{y^4a,y^4a^x\}$. So if $A\neq 1$, then $A\setminus 1 = \{a, a^x\}$ and $|A| = 3$, contradiction. If $A=1$, then $G$ is cyclic and $G=M$, contradiction.


