Standard Wreath Product and Sylow Subgroups A Sylow $p$-subgroup of $S_{p^r}$ is isomorphic with the standard Wreath Product $W(p,r) = (\cdots(C_p \wr C_p) \wr \cdots) \wr C_P)$, the number of factors being $r$.
I have a great doubt as to demonstrate this fact. Follow the solution proposed in Robinson's book (page 41/42) - A Couse in Group Theory
 A: One way to do it is to show that 


*

*$W(p, r)$ is, by construction, a permutation group of degree $p^{r}$, and then  

*the order of $W(p, r)$ is the $p$-part of $(p^{r})!$.

A: Let us prove that a Sylow $p$-subgroup of $S_{p^r}$ is of the required type by induction on $r$. The case $r=0$ is obvious. Assume that $S_{p^r}$ has as Sylow $p$-subgroup $P$ of the sort described. Consider the permutation
$$\pi = (1,1+P^r, \ldots, 1+(p-1)p^r)(2,2+p^r, \ldots, 2+(p-1)p^r) \cdots (p^r,p^r+p^r, \ldots, p^r+(p-1)p^r). $$
Then $\pi \in S_{p^{r+1}}$ and $\pi^p = 1$. Let $S_{p^r}$ be regarded as a subgroup of $S_{p^{r+1}}$  through its action on $\{1,2, \ldots, p^r\}$, the other symbols being fixed. Then $P_i = \pi^{-i}P\pi^i$ affects only the symbols $j+ip^r$, where $j=1,2, \ldots, p^r$. Hence, $P=P_0, P_1, \ldots, P_{p-1}$ generate their direct product; also $\pi^{-1}P_i\pi = P_{i+1}, 0 \leq i < p-1$, and $\pi^{-1}P_{p-1}\pi = P_{0}$. Thus $<\pi, P> \cong P \wr <\pi>$, the standard wreath product. It follows that $<\pi, P>$ is a Sylow $p$-subgroup of $S_{p^{r+1}}$.
A: If you want a reference to a good explanation, then I would recommend Dixon and Mortimer's explanation in Example 2.6.1 on pages 48 and 49 of their book Permutation Groups.
