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Does anyone know of a homomorphism from a group $G$ to another group with kernel as the intersection of all Sylow $p$-subgroups?

I was trying to prove that the intersection of Sylow subgroups is normal like this but couldn't think of one. I ended up showing it was characteristic instead. I'm curious if there is a easy homomorphism with this kernel though.

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    $\begingroup$ You could use that every conjugate of a Sylow $p$-subgroup is another Sylow $p$-subgroup so that $g(\bigcap P_i )g^{-1}=\bigcap (gP_ig^{-1})=\bigcap P_i$. $\endgroup$ – Taylor Jul 28 '15 at 22:50
  • $\begingroup$ I don't think you can necessarily apply the conjugacy condition here. For instance, if there are, say, 3 sylow subgroups P,Q,R and $g$ is chosen such that $gPg^{−1}=gQg^{−1}=gRg^{−1}=P$. Then $\bigcap gP_{i}g^{−1}=P$. The conjugacy condition says that there exists g∈G such that the Sylow subgroups are conjugate, but not necessarily for a fixed g∈G is every Sylow subgroup obtained by conjugating the Sylow subgroups by the fixed $g$. $\endgroup$ – TuoTuo Jul 28 '15 at 23:28
  • $\begingroup$ @Andrew Just cancel out the $g$'s and you will see that $P=Q=R$. $\endgroup$ – zibadawa timmy Jul 28 '15 at 23:33
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    $\begingroup$ @Andrew I understand your reservations; perhaps this will help: for any Sylow $p$-sugroup $P$ and $g\in G$, $gPg^{-1}<Q$ for some Sylow $p$-subgroup $Q$. Then $P<g^{-1}Qg$ and $g^{-1}Qg$ is necessarily a $p$-subgroup so that the maximality of $P$ implies $P=g^{-1}Qg$. $\endgroup$ – Taylor Jul 29 '15 at 3:54
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Well, all characteristic subgroups are necessarily normal, so that suffices. As Taylor's comment points out, using that Sylow subgroups are conjugate you know that $g\left(\bigcap P_i\right)g^{-1}=\bigcap g P_i g^{-1}$, which is an intersection of Sylow subgroups. Indeed, we necessarily have $\bigcap g P_i g^{-1}= \bigcap P_i$: if $g P_i g^{-1} = g P_j g^{-1}$, then $P_i=P_j$, and given any sylow subgroup $P$ then $Q=g^{-1}P g$ is also a Sylow subgroup and $gQg^{-1}=P$.

You can then construct a homomorphism with the appropriate kernel via the coset map.

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  • $\begingroup$ I guess this is essentially the same as my proof using characteristic. $\endgroup$ – TuoTuo Jul 29 '15 at 0:10
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In addition to the already given answers: $$\cap_{P \in Syl_p(G)}P= core_G(P)=O_p(G)$$ is the unique largest normal $p$-subgroup of $G$. It plays an important role in the theory of groups that are ($p$-)solvable.

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