Sylow theorem Cyclic sylow p subgroup $p$ is the smallest prime dividing order of $G$. $P$ is a sylow p subgroup which is cyclic. Prove that $N_G(P) = C_G(P)$
This is my approach :
Since $P$ is sylow p subgroup so its order is some power of $p$. Now $N_G(P)/C_G(P)$ is isomorphic to a subgroup of $Aut(P)$ which has order $p^{(a-1)}(p-1)$. Now If $p$ doesn't divide index of $C(P)$ in $N(P)$ I'm through, but I'm not being able to proof why $p$ will not divide $[N(P):C(P)]$
 A: First consider the case when $C_G(P)$ is a $p$-subggroup of $G$.
Now since $C_G(P)\subseteq N_G(P)$, we have $C_G(P)\subseteq P$. (we have used the fact that $C_G(P)$ is a $p$-subgroup here.)
But since $P$ is cyclic we also have $P\subseteq C_G(P)$.
Thus $C_G(P)=P$.
Now by the $N/C$ theorem, we have $|N_G(P)/C_G(P)|$ divides $|\text{Aut}(P)|=p^{n-1}(p-1)$.
Using $C_G(P)=P$, we have $|N_G(P)|$ divides $p^{2n-1}(p-1)$.
Here use the fact that $p$ is the smallest prime dividing $|G|$ to deduce that $|N_G(P)|$ is a $p$-subgroup of $G$.
Since $P\subseteq N_G(P)$, we must have $|N_G(P)|=p^n$, and therefore $N_G(P)=P$.
This gives $N_G(P)=C_G(P)$.
Now suppose that $C_G(p)$ is not a $p$-subgroup of $G$.
Then, since $P$ is cyclic, $P\subsetneq C_G(P)$.
Using the $N/C$ theorem we have $|N_G(P)/C_G(P)|$ divides $p^{n-1}(p-1)$.
But note that $p$ does not divide $|N_G(P)/C_G(P)|$.
Say $|N_G(P)/C_G(P)|=k$.
Then $k|[p^{n-1}(p-1)]$.
Suppose $k>1$.
Since any prime factor of $k$ is greater than $p$, we have $\gcd(k,p)=1$.
Therefore $k|(p-1)$ but this is not possible since any prime factor if $k$ is greater than $p$.
Therefore $k=1$ and we are done.
