$P \cap N$ is a Sylow $p$-subgroup of $N$, where $N$ is normal in $G$ and $P$ is a Sylow $p$-subgroup of $G$? In 'A Course in Group Theory' by Humphreys, Proposition 11.14 says that if $G$ is a finite group, $P$ is a Sylow $p$-subgroup of $G$ and $N$ is a normal subgroup of $G$, then $P \cap N$ is a Sylow $p$-subgroup of $N$.
But doesn't this fail in the case where $P \cap N = \{1\}$? I thought the definition of a Sylow $p$-subgroup excludes $\lbrace1 \rbrace$, since the definition of Sylow $p$-subgroup requires an order $p^n$ where $n$ is a positive integer (i.e. not zero). If not, we need to adjust the definition of Sylow $p$-subgroup to a subgroup of order $p^n$ where $n$ is non-negative. However I don't think Humphreys intends this, as in his answer to exercise 11.3 he uses $\{1\}$ as an example of a group that is not a Sylow 2-subgroup.
As an example: I think we could have $G = D_3$ (the dihedral group with 6 elements), $N = \langle a \rangle$ where $a$ is a rotation, and $P = \langle b \rangle$ where $b$ is a reflection. Then $P \cap N = \{1\}$, which is not a Sylow 2-subgroup of $N$ although $N$ is normal in $G$ and $P$ is a Sylow 2-subgroup. Please let me know what you think. Thanks

This may be a bit unsatisfying, but the answer is that the Sylow $p$-subgroup of a finite group $G$ for which $|G|$ is not divisible by $p$ is simply the identity.  Note that $1$ is indeed a power of $p$ for any prime $p$, so if $p^0=1$ is the highest power of $p$ dividing $|G|$, then $\{1\}$ is the Sylow $p$-subgroup.  So, $P\cap N=1$ is not a contradition because $P$ is a Sylow $2$-subgroup of $D_3$ and $\{1\}$ is, in fact, the Sylow $2$-subgroup of $\langle a \rangle \cong C_3$.  (For an even easier example, consider $\mathbb{Z}_6$.)
If the book doesn't define this way, it's possible that the writer just forgot to write an exception to this Proposition.  This is standard in group theory literature.
