group with $1+rp$ Sylow $p$-subgroups and existence of group in $\mathrm{Sym}(1+rp)$ with $1+rp$ Sylow $p$-subgroups Let $r\in \mathbb{N}$ and $p$ a prime. Suppose that a group $G$ has $1+rp$ Sylow $p$-subgroups. Then there exists $H\leq \mathrm{Sym}(1+rp)$ that has precisely $1+rp$ Sylow $p$-subgroups.
I was thinking about Cayley's theorem which says that every group is isomorphic to a group of permutations, so $G$ would embed in some $\mathrm{Sym}(n)$, but not sure precisely why it would be in $\mathrm{Sym}(1+rp)$.
 A: Theorem Let $G$ be a finite group having exactly $n$ Sylow $p$-subgroups for some prime $p$. Then there exists a subgroup $H$ of the symmetric group $S_n$ such that $H$ also has exactly $n$ Sylow $p$-subgroups. 
Proof Let $P$ be a Sylow $p$-subgroup of $G$ and set $N = N_G(P)$. By assumption,$ |G : N| = n$ and therefore the $n!$-Theorem (see Theorem 1.1. here) implies that there is a homomorphism $\bar{ }: G \rightarrow S_n$ with kernel $K \subseteq N$. We know that $\overline{PK} = PK/K$ is a Sylow $p$-subgroup of $\overline{G} ⊆ S_n$.
Furthermore, put $M = N_G(PK)$, then the one-to-one correspondence between the subgroups of $\overline{G}$ and the subgroups of $G$ containing $K$ implies that $\overline{M} = N_\overline{G}(\overline{PK})$ and that $|\overline{G} : \overline{M}| = |G : M|$. Let us identify $M$. First, since $K \unlhd G$, we have $N=N_G(P) \subseteq M$. Conversely, since $PK \unlhd M$, the Frattini argument implies that $M = N_M(P)PK \subseteq N_G(P)=N$. So $M=N$ and $|\overline{G}:\overline{M}| = |G : M| = |G : N| = n$.
