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I am aiming for looking for all the Sylow subgroups of classical groups, which gives me a seemingly elementary question: what can we say about Sylow subgroups of a subgroup if we know all Sylow subgroup of the original one?

In my case, I know some Sylow subgroups of $GL(n,F_q)$ in which $F_q$ is a finite field and I wonder can I find some Sylow subgroups of $SL(n,F_q)$ from those and vice versa.

I found this post Counterexample of Sylow subgroups of a subgroup and if the writer is correct, then let $P$ be a Sylow subgroup of $GL$, then $P\cap SL$ is a Sylow subgroup of $SL$ since $SL$ is a normal subgroup of $GL$.

This is my question, given a Sylow subgroup $S$ of $SL$, then is there a Sylow subgroup $P$ of $GL$ s.t. $S=P\cap SL$?

I looked for examples but failed. However, if this is true, then there would be a very beautiful relation between Sylow subgroups of $SL$ and $GL$, which is too good to be true.

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    $\begingroup$ The answer to your question is yes because ${\rm SL}(n,K)$ is a normal subgroup of ${\rm GL}(n,K)$ for all fields $K$. $\endgroup$ – Derek Holt Jun 23 at 6:53
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If $H \leq G$, then a $p$-subgroup of $H$ is also a $p$-subgroup of $G$, and so is contained in a Sylow $p$-subgroup of $G$, and every $p$-subgroup is contained in a Sylow $p$-subgroup. In other words, yes, a Sylow $p$-subgroup of $H$ must be contained in a Sylow $p$-subgroup of $G$.

In the case that $H \trianglelefteq G$, then since all Sylow $p$-subgroups of any group are conjugate (for a fixed $p$), and $H^{g} = H$ (conjugation of $H$ by $g$) for all $g \in G$, then we also have conversely that every Sylow $p$-subgroup of $G$ meets $H$ in a Sylow $p$-subgroup.

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