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I'll state the result I'm trying to prove, progress I've made, and the two questions I have which will help me solve it. The question is originally motivated by studying defect groups in modular representation theory.

Let $D$ be be a subgroup of a finite group $G$ with the following property:

If $S$ is a Sylow $p$-subgroup of $G$ containing $D$ then there is $c\in C(D)$ such that $D=S\cap cSc^{-1}$.

Prove that $D$ is the largest normal $p$-subgroup of $N_G(D)$.

The start of my attempt at a proof: It will suffice to show that $D$ is the intersection of Sylow $p$-subgroups of $N_G(D)$, since every normal $p$-subgroup of $N_G(D)$ is contained in every Sylow $p$-subgroup of $N_G(D)$, by Sylow II. We do not however know that $cSc^{-1}$ is Sylow (Sylow II states that all Sylows are conjugate but not necessarily that all conjugates are Sylow; right?)

Questions:

Q1: If we let $T$ be a Sylow $p$-subgroup of $N_G(D)$, can we necessarily find a Sylow $p$-subgroup $S$ of $G$ containing $T$?

Q2: If $T$ is a Sylow $p$-subgroup of $N_G(D)$, must $T\supseteq D$ and $cTc^{-1}\supseteq D$ for any $c\in C(D)$?

(If the answer to these two is yes then the result will follow).

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Presumably you want $D$ to be a $p$-subgroup. Now take a Sylow $p$-subgroup $T$ of $N_G(D)$; then $D$ is contained in $T$ and $T$ is contained in some Sylow $p$-subgroup $S$ of $G$. Now there exists, by hypothesis, $c\in C_G(D)$ such that $D=S\cap S^c$. Since $c\in N_G(D)$, $D=S\cap S^c\cap T = T\cap S^c$, so that $T\cap T^c\le D$; this of course implies $D$ is the largest normal p-subgroup of $N_G(D)$. –  user641 Sep 7 '11 at 23:53
    
Why is it important that $c\in N_G(D)$ to see that $D=S\cap S^c\cap T = T\cap S^c \le T\cap T^c$ and why does this imply $D$ is the largest normal $p$-subgroup of $N_G(D)$? –  Clinton Boys Sep 8 '11 at 0:44
    
It is important because then $T^c$ is also a Sylow of $N_G(D)$. So if I let $O_p$ denote the intersection of all Sylows in $N_G(D)$, then $O_p\le T\cap T^c\le D$, and since $D$ is normal in $N_G(D)$, we actually have $O_p=D$. –  user641 Sep 8 '11 at 0:50
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Ah right. We don't need $c\in N_G(D)$ to see $D\geq T\cap T^c$, but for the other direction, $D\leq T$ so $D\leq T^c$ since $c\in N_G(D)$. Hence $D=T\cap T^c$ is the intersection of Sylow $p$-subgroups of $N_G(D)$, whence is the largest normal $p$-subgroup, since every normal $p$-subgroup is contained in every Sylow $p$-subgroup. –  Clinton Boys Sep 8 '11 at 0:52
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1 Answer

up vote 1 down vote accepted

It is true that a conjugate of a Sylow subgroup is still a Sylow subgroup, because conjugation is an automorphism. Onward to the questions.

For Q1, note that $T$ is in particular a $p$-subgroup of $G$. Every such subgroup is contained in a $p$-Sylow subgroup of $G$. See Theorem 5.6(c) of Milne's notes for a typical proof of this.

I think Q2 is true as well. Since $D$ is a $p$-subgroup of $N_G(D)$, it is certainly contained in some $p$-Sylow subgroup $P$ of $N_G(D)$ by the theorem mentioned above. By Sylow, there is an element $x \in N_G(D)$ such that $xPx^{-1} = T$, and of course $xPx^{-1} \supset xDx^{-1} = D$. After writing that last containment down it should be clear that $cTc^{-1} \supset D$ for $c \in C_G(D)$.

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By $C(D)$ I mean the centralizer of $D$ in $G$. –  Clinton Boys Sep 7 '11 at 23:59
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