Counterexample of Sylow subgroups of a subgroup Let $P$ be a Sylow subgroup of a finite group $G$. Let $N$ be a subgroup of $G$.
$(1)$ If $N$ is normal in $G$, then $P\cap N$ is a Sylow subgroup  of $N$. I have proved this.
$(2)$ In general, is $P\cap N$ a Sylow subgroup of $N$? I think this is not right since the proof does not hold. How to give a counterexample? I am quite confused. Thanks.
 A: Tobias and Nicky already answered your question, but we can actually classify the counterexamples amongst finite groups.
Classification
Kegel (1962) tried to understand exactly those subgroups $N$ that work. He called such subgroups $p$-subnormal if it worked for all the Sylow $p$-subgroups.
Definition: Let $G$ be a group, and $N \leq G$ a subgroup. 


*

*$N$ is said to be $p$-subnormal in $G$ iff $P \cap N$ is a Sylow $p$-subgroup of $N$ for every Sylow $p$-subgroup $P$ of $G$.

*$N$ is said to be subnormal in $G$ iff there is a sequence of subgroups $N = N_0 \unlhd N_1 \unlhd \ldots \unlhd N_k = G$ each normal in the next, starting at $N$ and ending at $G$


The Kegel–Wielandt conjecture was that a subgroup was $p$-subnormal for all primes $p$ iff the subgroup was subnormal. Kleidman (1991) proved this:
Theorem: If $G$ is a finite group and $N \leq G$, then the following are equivalent:


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*$N$ is $p$-subnormal in $G$ for all primes $p$

*$N$ is subnormal
$2 \implies 1$ is easy to prove form your (1), and $1 \implies 2$ points exactly where counterexamples occur: amongst finite groups, exactly in the non-nilpotent groups.
Smallest example
The smallest non-nilpotent group is the non-abelian group of order 6. It has more than one Sylow 2-subgroup, for instance $N$ and $P$. Their intersection is $P \cap N = 1$ and is too small to be a Sylow 2-subgroup of $N$. This is Tobias's example. Nicky's example is less trivial.
Bibliography


*

*Kegel, Otto H.
“Sylow-Gruppen und Subnormalteiler endlicher Gruppen.”
Math. Z. 78 (1962) 205–221.
MR147527
DOI:10.1007/BF01195169

*Wielandt, Helmut.
“Zusammengesetzte Gruppen: Hölders Programm heute.”
The Santa Cruz Conference on Finite Groups (Univ. California, Santa Cruz, Calif., 1979),
pp. 161–173, Proc. Sympos. Pure Math., 37, Amer. Math. Soc., Providence, R.I., 1980.
MR604575

*Kleidman, Peter B.
“A proof of the Kegel-Wielandt conjecture on subnormal subgroups.”
Ann. of Math. (2) 133 (1991), no. 2, 369–428.
MR1097243
DOI:10.2307/2944342
A: If $H$ is a subgroup of $G$ and $S \in Syl_p(H)$, then $S=H \cap P$ for some $P \in Syl_p(G)$. But the converse is not true: try to find the Sylow 2-subgroups of $S_3 \times S_3$ and intersect these with (a copy of) its subgroup $C_2 \times S_3$ of order 12.
