Let $P$ be a $p$-Sylow subgroup of a finite group $G$, and suppose $a, b$ are in the center of $P$. Suppose further that there is an $x$ in $G$ such that $a = xbx^{-1}$. Then there exists $y$ in $N(P)$ with $a = yby^{-1}$, where $N(P)$ is the normalizer of $P$.

I've been working on this problem since last night and have made basically no progress. Any ideas?


Hint: Show that both $P$ and $xPx^{-1}$ are subgroups of $C_G(a)$, the centralizer of $a$ in $G$, and hence are $p$-Sylow subgroups of $C_G(a)$.

  • $\begingroup$ Thanks for the hint, but I don't really know where to go from here. How do I relate this group to N(P)? $\endgroup$ – Vik78 Aug 29 '16 at 19:24
  • $\begingroup$ Okay, so then $P$ and $xPx^{-1}$ are conjugate in the centralizer, so there exists $y$ in the centralizer with $P = yxPx^{-1}y^{-1}$. Therefore $yx$ is in the normalizer, but since $y$ commutes with $a$ we have $a = xbx^{-1} = yxbx^{-1}y^{-1}$. Many thanks! $\endgroup$ – Vik78 Aug 29 '16 at 19:32

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