# Show that $P$ is a Sylow $p$-subgroup of $G$ if and only if $P$ is a Sylow $p$-subgroup of $N_G(P).$

Question: Let $$G$$ be a finite group and $$P$$ be a $$p$$-subgroup of $$G.$$ Show that $$P$$ is a Sylow $$p$$-subgroup of $$G$$ if and only if $$P$$ is a Sylow $$p$$-subgroup of $$N_G(P).$$

My attempt: Suppose that $$|P| = p^r$$

$$(\Rightarrow)$$ Assume that $$|G| = p^rm$$ where $$r$$ does not divide $$m.$$ Since $$P\leq N_G(P)\leq G,$$ by Lagrange's Theorem, we have $$p^r$$ divides $$|N_G(P)|$$ and $$|N_G(P)|$$ divides $$p^rm.$$ It follows that $$|N_G(P)| = p^r n$$ where $$p$$ does not divide $$n.$$ Therefore, $$P$$ is a Sylow $$p$$-subgroup of $$|N_G(P)|.$$

$$(\Leftarrow):$$ Suppose that $$|N_G(P)| = p^r n$$ where $$p$$ doe not divide $$n.$$ I do not know how to proceed from here.

Any hint would be appreciated.

• Your first implication is really an "if and only if". A subgroup is defined to be a $p$-Sylow subgroup if it is a group of order $p^r$, where $p^r$ is the largest power of $p$ dividing $|G|$. Since $P \leq N_G(P) \leq G$, you have that $p^r$ is the largest power of $p$ dividing both $|G|$ and $|N_G(P)|$. – Joe Johnson 126 Oct 5 '18 at 0:35
• @JoeJohnson126 I understand that $p^r$ is the largest power of $p$ dividing $|N_G(P).|$ But I do not see how does $p^r$ is the largest power of $p$ divide $|G|.$ – Idonknow Oct 5 '18 at 1:04
• For the other direction you need two things: 1. Any $p$-subgroup is contained in a Sylow $p$-subgroup. 2. In a $p$-group, all proper subgroups are properly contained in their normalizer. – Tobias Kildetoft Oct 5 '18 at 3:51
• You may want to add an assumption that $G$ is finite. It might still be true otherwise, for all I know, but there may be more to the proof than intended. – C Monsour Oct 5 '18 at 17:42
• @CMonsour Edited. Thanks. – Idonknow Oct 7 '18 at 2:41

The answer to the question follows immediately from the following. In general, if $$P$$ is a $$p$$-subgroup of $$G$$ (so not necessarily Sylow) then $$|G:P| \equiv |N_G(P):P|$$ mod $$p$$. See for instance http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/sylowpf.pdf for a proof. The proof is not difficult and depends on the action of $$P$$ on the left coset space $$G/P$$ by left multplication.
• Yes, I manage to prove that $|G:P| \cong |N_G(P):P|$ if $P$ is a $p$-subgroup of $G.$ However, I do not see how the result can solve my problem. Any hint would be appreciated. – Idonknow Oct 7 '18 at 2:36
• Assume that $P$ is a Sylow $p$-subgroup of $N_G(P)$, but not of $G$. Then $[N_G(P) : P]$ is not divisible by $p$, while $[G : P]$ is. But this is impossible, since $[G : P] \equiv [N_G(P) : P] \pmod{p}$. – D_S Oct 7 '18 at 4:20