Let $G$ be a finite group having exactly $n$ Sylow $p$-subgroups for some prime $p$. Show that there exists a subgroup $H$ of the symmetric group $S_n$ such that $H$ also has exactly $n$ Sylow $p$-subgroups.

My attempt: Say $G$ has only $1$ Sylow $p$-subgroups for some $p$ ($n=1$). Then, $S_n=S_1=\{e\}$ so it has no subgroup. Couldn't it be a counterexample? Otherwise, I should look into the case $n>p$, because $n\equiv 1\ (mod\ p)\implies n=1\ or\ n>p$ by the 2nd Sylow theorem.

In that case ($n>p$), $p$ devides $|S_n|=n!$, so $S_n$ has Sylow $p$-subgroups by the 1st Sylow theorem. Say $P_1,P_2,...,P_k$ are them. Now by the 3rd Sylow theorem, $k=(S_n:N_{S_n}(P_i))$ for $i=1,2,...,k.$ And from here, I want to show that $k=n$ and $N_{S_n}(P_i) = N_{S_n}(P_j)$ for $i,j=1,2,...,k$ and conclude $\exists H=N_{S_n}(P_1)$, it seems gone wrong.

So I have two main troubles here: Can't discard the case $n=1$, and can't go further beyond $p|n!$ in the $n>p$ case (I guess my direction - trying to show $H=N_{S_n}(P_1)$ is just wrong?).

Any hints or suggestions would be appreciated.

  • 1
    $\begingroup$ The case $n=1$ is not a counterexample, because the trivial subgroup is a Sylow $p$-subgroup of $S_1$ for all primes $p$. For the general case, let $S$ be the set of $n$ Sylow $p$-subgroups of $G$. Then $G$ acts on $S$ by conjugation, and the image of $G$ in ${\rm Sym}(S) \cong S_n$ defined by this action has exactly $n$ Sylow $p$-subgroups. $\endgroup$ – Derek Holt Feb 11 at 9:09

An equivalent action to the action given by @Nicky is the action of $G$ on the conjugates of $P$ for some Sylow $p$-subgroup of $G$ defined by conjugation in $G$ (so for $x,y\in G$ $g\cdot xPx^{-1}=gxPx^{-1}g^{-1}$). I'll skip the details as @Nicky's answers seems complete.

As there are $n$ Sylow subgroups of $G$ (precisely the conjugates of $P$), this defines a homomorphism $f:G\to S_n$.

Note that $f(G)\cong G/K$ where $K=\ker(f)$ and $f(G)\le S_n$. So $f(H)$ is a Sylow $p$-subgroup of $f(G)$. The action of $f(G)$ on the conjugates of $f(H)$ is equivalent to the action of $G$ on the conjugates of $H$ so $f(G)$, so $f(H)$ must have $n$ conjugates. That is $f(G)$ has $n$ Sylow $p$-subgroups.


Let $P \in Syl_p(G)$ and hence $|G:N_G(P)|=n$. The $n!$-Theorem ($G$ acting on the right cosets of in this case $N_G(P)$ by right multiplication, see M. Isaacs, Finite Group Theory Theorem (1.1) for example) implies that there is a homomorphism $\bar{} : G \rightarrow S_n$, with kernel $K=core_G(N_G(P))$. We know that $\overline{PK}=PK/K$ is a Sylow $p$-subgroup of $\bar{G}$ a subgroup of $S_n$.

Observe that there is a one-to-one correspondence between the subgroups of $\overline{G}$ and the the subgroups of $G$ containing $K$. This implies that $\overline{N_G(PK)}=N_{\overline{G}}(\overline{PK})$ and in particular $|\overline{G}:N_{\overline{G}}(\overline{PK})|=|G:N_G(PK)|$.

Finally we will show that in fact $N_G(PK)=N_G(P)$ and then we are done. Since $K \lhd G$ it is easy to see that $N_G(P) \subseteq N_G(PK)$. But then, since $PK \lhd N_G(PK)$ and $P \in Syl_p(PK)$, the Frattini argument gives $N_G(PK)=N_{N_G(PK)}(P)PK=(N_G(P) \cap N_G(PK))PK=N_G(P)PK=N_G(P)$.


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