Just as a note, please don't give me the proof, I feel like I'm close and am just looking for a point in the right direction.
I am trying to prove that if $G$ is a finite group with $|G|$ = $p^nm$, and $N \lhd G$, such that $p||N|$, and $P \lt G$ is a sylow p-subgroup then $P \cap N$ is also a sylow p-subgroup of $N$.
I have so far that there is a sylow p-subgroup in N because $p||N|$, and now I am trying to use Sylow's 2nd Theorem along with the fact that $N \lhd G$ and that all sylow subgroups are conjugate to each other.
My proof so far is as follows:
$P_N \lt N \lhd G \Rightarrow P_N \lt G$ so $P_N$ is also a slyow p subgroup of G,
$\Rightarrow g(P_N)g^{-1} = P$ by the second sylow theorem $\Rightarrow P\cap N = P_N$
$\therefore P \cap N$ is a sylow p-subgroup $\blacksquare$
I'm not sure where the flaw in my reasoning is, but I don't use the fact that $N$ is normal at all, which is my first hint that I'm wrong, so my idea is to use the second theorem again somehow between finding $P_N$ in $N$ and claiming that $P_N \lt G$. Any advice is greatly appreciated, thank you.
I apologize for confusion, I am not good at English lol, but I think I fixed my problems
Let $G=Z_{12}, P_N=N=\{0, 6\}, $, but $P_N$ is not Sylow p-subgroup of $G$.