Let $|G|=14924=2^2 \times 7 \times 13 \times 41.$ We can see Sylow $41$-subgroup is normal But when I try for Sylow $7$ and $13$, I face some problems. I use the argument following:
Let $P$ be a Sylow $41$ and $R$ be a Sylow $7$ subgroup. $NG(P)/CG(P)$ can be imbedded $Aut(P)$ and since $P$ is normal $|G|/CG(P)$ can be imbedded $C_{40}$. If $|G|/CG(P)$ only possible $1$, I am done. But since $G$ contains factor $2$, I cannot obtain $1$. İt can be $1$,$2$ or $4$. How Can I continue to show that Sylow $7$ and $13$ is normal?
I'll answer this (old, very old) question since it's moved up the queue, is unanswered, and it's not a usual element-counting one. The Sylow $41$-subgroup is normal, and $G/C_G(P)$ is a $2$-group. Thus $P$ is centralized by an element of order $7$ and one of order $13$. The number of Sylow $7$-subgroups $R$ is $1$ or $533$, but $41\mid 533$ and we already know that $41\mid |C_G(R)$. Thus $n_7=1$. In particular, $G/C_G(R)$ is a $2$-group as well. This means that an element of order $13$ centralizes $R$. Hence if $Q$ is a Sylow $13$-subgroup, we already know that $|C_G(Q)|$ is divisible by $7\cdot 13\cdot 41$, so has index at most $4$. In particular, $n_{13}=1$, as needed.