# If a $p$-Sylow subgroup is in the normalizer of another, they're equal?

I've gotten a little tripped on part of what I'm guessing is a straight forward problem. This is part (a) of Exercise 23 in Lang's Algebra.

Let $P,P'$ be $p$-Sylow subgroups of a finite group $G$. If $P'\subset N(P)$ (normalizer of $P$), then $P'=P$.

How can I go about this? I know that $P'=gPg^{-1}$ for some $g\in G$, and I figure conjugation must come into play if the normalizer is part of the question. I also figure it'd be enough to show one is contained in the other since they're both maximal $p$-subgroups. I haven't really been able to scratch out more than that.

I'd appreciate any hints or tips/tricks or answers to this part if you have them. Thanks.

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Hint: If a $p$-Sylow subgroup is normal, then it is the only $p$-Sylow subgroup of that group. By definition, $P$ is a normal subgroup of $N(P)$. Therefore...
Hm, so $P\unlhd N(P)$, but $P'\leq N(P)$ and a $p$-Sylow subgroup too, so $P=P'$. Thanks Ted and @Zev, you guys were both lightning fast. – yunone Sep 12 '11 at 5:31
• $N(P)$ is the largest subgroup of $G$ having $P$ as a normal subgroup.
• If $P$ is a Sylow $p$-subgroup of $G$, then it is also a Sylow $p$-subgroup of any subgroup $H\subset G$ containing it.