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Let $G$ be a finite solvable group and $N$ be a normal subgroup of $G$ such that $G/N$ is not abelian. Also for every prime integer $p$, $G$ has at most $5$ conjugacy classes whose sizes are multiples of $p$. Moreover, let $G/N$ be a Frobenius group with kernel $K/N$ of order $5$ and complement isomorphic to $G/K$ of order $2$. Let $N/L$ be a chief factor of $G$. Therefore $N/L$ is an elementary abelian $p$-group. If $|N/L|>2$, then can we say that $C_{G/L}(N/L)=N/L$?

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Let $C=C_{G/L}(N/L)$. Clearly, $N/L\leq C$, so suppose $N/L<C$. Then $K/L\leq C$ and thus $K/L$ is abelian. Since $K/L$ has index $2$ in $G/L$, the conjugacy classes (in $G/L$) of elements of $K/L$ have size either $1$ or $2$.

Since $G/N\cong D_5$, the center of $G/L$ is contained in $N/L$ and thus every element of $K/L\setminus N/L$ is in a conjugacy class (in $G/L$) of size exactly $2$. Since $|N/L|\geq 3$, that's at least $4|N/L|\geq 12$ elements, split into at least $6$ classes of size exactly $2$.

When we go to $G$, these will become at least $6$ conjugacy classes of even size, a contradiction.

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