I'm trying to solve the following problem:

"Let $P$ be a $p$–Sylow subgroup of a finite group $G$. Consider the set of conjugate subgroups $gPg^{-1}$ with $g \in G$. Show that the number of conjugates is not divisible by $p$."

My attempt:

the number of conjugates of a $p$–Sylow subgroup is congruent to $1$ modulo $p$. Let's assume that $p$ divides the number of conjugates, then it is false that the number of conjugates is congruent to $1$ modulo $p$, so $p$ does not divide the number of conjugates.

Is this correct? I'm not sure if all the conjugates of a $p$–Sylow subgroup are $p$–Sylow subgroups (the Second Sylow Theorem says that two $p$–Sylow subgroups are conjugate).


  • 9
    $\begingroup$ Your solution is correct, but you don't need to use the theorem that the number of conjugates is congruent to $1$ mod $p$. Then number of conjugates is $|G|/N_G(P)|$, and since $P \le N_G(P)$ and $p$ does not divide $|G|/|P|$, it cannot divide $|G|/|N_G(P)|$. $\endgroup$ – Derek Holt Apr 19 '16 at 1:57
  • 2
    $\begingroup$ It is easy to see that conjugates of Sylow $p$-subgroups are themselves Sylow $p$-subgroups: conjugate subgroups are isomorphic, and thus have the same order. $\endgroup$ – stochasm Apr 19 '16 at 3:15

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