$p$ 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 $[N(P):C(P)]$

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    $\begingroup$ Hint: $P\subseteq C(P)$. $\endgroup$ – Tobias Kildetoft Feb 7 '14 at 14:05
  • $\begingroup$ @user115654 No, that is not true. Take for example the cyclic subgroup of index $2$ in $Q_8$. This is normal but not central. $\endgroup$ – Tobias Kildetoft Feb 8 '14 at 9:22
  • $\begingroup$ @TobiasKildetoft: You're right. What I meant to say was: if $P$ is a cyclic subgroup, contained in the center of a Sylow $p$-subgroup, then $N(P) = C(P)$ (for $p$ the smallest prime dividing $|G|$). $\endgroup$ – zcn Feb 8 '14 at 9:31
  • $\begingroup$ @user115654 Indeed (still assuming $p$ is the smallest prime dividing the order of the group). In fact, in that case, the group has a normal subgroup of order coprime to $p$. $\endgroup$ – Tobias Kildetoft Feb 8 '14 at 9:34
  • $\begingroup$ @TobiasKildetoft: Your last statement cannot be true (except in a trivial case). Sylow subgroups for the smallest prime always exist, and their centers always contain nontrivial cyclic subgroups, but the whole group need not have a normal subgroup of order coprime to $p$ besides $\{e\}$ (e.g. if $G$ is simple) $\endgroup$ – zcn Feb 8 '14 at 9:46

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.

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  • $\begingroup$ I'm having a tough time with this step "Here use the fact that p is the smallest prime dividing |G| to deduce that |NG(P)| is a p-subgroup of G. Since P⊆NG(P), we must have |NG(P)|=pn, and therefore NG(P)=P." Could you elaborate how exactly we know NG(P) is a p-subgroup? $\endgroup$ – Algebra is Awesome Oct 10 '19 at 1:15
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    $\begingroup$ As shown in the previous line, we have that $|N_G(P)|$ divides $p^{2n-1}(p-1)$. If $N_G(P)$ were not a $p$-group then there would be a prime $q\neq p$ such that $q|N_G(P)$. But since $p$ is the smallest prime dividing $|G|$, we have that $q>p$. But then $q$ divides $p^{2n-1}(p-1)$, giving that $q|(p-1)$. But then $q<p$ and we have a contradiction. $\endgroup$ – caffeinemachine Oct 10 '19 at 9:53

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