Any Sylow $p$-subgroup of $GL_2(p)$ is a Sylow $p$-subgroup of $SL_2(p)$. Is this true? I know that a Sylow $p$-subgroup of $GL_2(p)$ has order $p$ and so does a Sylow $p$-subgroup of $SL_2(p)$. Does this mean that $Syl_p(GL_2(p)) = Syl_p(SL_2(p))$?
 A: More generally if $N \unlhd G$ and $[G:N]$ is relatively prime to $p$, then the Sylow $p$-subgroups of $N$ and $G$ are the same. This handles your case, since $\operatorname{SL}_2(p) \unlhd \operatorname{GL}_2(p)$ (being the kernel of the determinant homomorphism) and $[\operatorname{GL}_2(p):\operatorname{SL}_2(p)]=p-1$ (from the first isomorphism theorem) is relatively prime to $p$.
Proof of the generalization:
If $P$ is a Sylow $p$-subgroup of $N$, then $P$ is also a Sylow $p$-subgroup of $G$ just because $[G:N]$ is relatively prime to $p$ (so the same power of $p$ divides both $|G|$ and $|B|$). If $Q$ is a Sylow $p$-subgroup of $G$, then Sylow's theorem says $P$ and $Q$ are conjugate, so $P^g = Q$ for some $g \in G$. Then $Q = P^g \leq N^g = N$ as long as $N \unlhd G$. Hence every Sylow $p$-subgroup of $N$ is a Sylow $p$-subgroup of $G$, and every Sylow $p$-subgroup of $G$ is a Sylow $p$-subgroup of $N$. $\square$
Smallest such $N$:
There is a unique smallest such normal subgroup $N$. It is called $O^{p'}(G)$ and is the intersection of all normal subgroups whose index is relatively prime to $p$. It is also the subgroup generated by the Sylow $p$-subgroups of $G$.
Counterexample:
If you don't require $N$ to be normal, then there may be Sylow $p$-subgroups of $G$ that are not Sylow $p$-subgroups of $N$. The smallest example is $G=S_3$, a non-abelian group of order $6$. Take $H=S_2=\langle (1,2) \rangle$ and $p=2$. Then $|G|$ and $|H|$ are divisible by the same power of $p$, so their Sylow $p$-subgroups have the same size. Every Sylow $p$-subgroup of $H$ is therefore a Sylow $p$-subgroup of $G$ (there is only one, $H$ itself). However, $G$ has two other Sylow $p$-subgroups: $\langle (1,3) \rangle$ and $\langle(2,3) \rangle$, neither of which is a Sylow $p$-subgroup of $H$ (because they are not even subgroups of $H$).
A: Recall that $|SL_2(p)|=p(p^2-1)$ and $|GL_2(p)|=p(p^2-1)(p-1)$. 
Any Sylow $p$-subgroup of $SL_2(p)$ has order $p$ and, hence, is a maximal $p$-subgroup of $GL_2(p)$.
On the other hand, $UT_2(p)$ is a Sylow $p$-subgroup of $GL_2(p)$. Then (i) $UT_2(p) < SL_2(p)$; and (ii) $S^{-1}UT_2(p)S < SL_2(p)$ for any $S \in GL_2(p)$. Thus, a Sylow $p$-subgroup of $GL_2(p)$ is also a Sylow $p$-subgroup in $SL_2(p)$.
Update: Here $UT_2(p)$ is the group of upper triangular matrices, s.t. all elements on the main diagonal are 1.
