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I'm trying to understand a classification of all Sylow $p$ subgroups of $S_n$.

Let $Z_p$ be the subgroup of $S_p$ generated by $(12\cdots p)$. Then $Z_p\wr Z_p$ has order $p^p\cdot p=p^{p+1}$, and is isomorphic to a subgroup of $S_{p^2}$.

Define inductively $Z_p^{\wr r}$ by $Z_p^{\wr 1}=Z_p$ and $Z_p^{\wr k+1}=Z_p^{\wr k}\wr Z_p$. It is easy to show by induction that $Z_p^{\wr r}$ has order $p^{(p^{r-1}+p^{r-2}+\cdots+1)}$, and since by inductively assuming $Z_p^{\wr r-1}$ is isomorphic to a subgroup of $S_{p^{r-1}}$ and $Z_p$ isomorphic to a subgroup of $S_p$, then $Z_p^{\wr r}$ is isomorphic to a subgroup of $S_{p^r}$.

However, I can't make the jump that if $n=a_0+a_1p+\cdots+a_kp^k$ is the base $p$ expansion, then any Sylow $p$-subgroup is isomorphic to $$ \underbrace{Z_p^{\wr 1}\times\cdots\times Z_p^{\wr 1}}_{a_1}\times \underbrace{Z_p^{\wr 2}\times\cdots\times Z_p^{\wr 2}}_{a_2}\times\cdots\times \underbrace{Z_p^{\wr k}\times\cdots\times Z_p^{\wr k}}_{a_k}. $$

I know this group has order $$ (p)^{a_1}(p^{p+1})^{a_2}\cdots(p^{(p^{k-1}+p^{k-2}+\cdots+1)})^{a_k}=p^{\sum_{i=1}^k a_i(1+\cdots+p^{i-1})}=p^{\nu_p(n!)} $$ which is the order of any Sylow $p$-subgroup of $S_n$, based on the formula here. However, I couldn't find an epimorphism from any Sylow $p$-subgroup onto this product, or vice versa. Is it clear how this is isomorphic to a subgroup of $S_n$? Then I understand that the isomorphism would just follow from the Sylow theorems. Thanks.

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This follows from the second Sylow theorem: all Sylow $p$-subgroups of a finite group are conjugate. – Qiaochu Yuan Jun 8 '12 at 20:55
Thanks, that's what I was thinking, but I'm having trouble seeing why the product is isomorphic to a subgroup of $S_n$ in the first place from the induction hypothesis. – Nastassja Jun 8 '12 at 20:58
There are embeddings $S_n \times S_m \to S_{n+m}$ for every $n, m \ge 0$. – Qiaochu Yuan Jun 8 '12 at 20:59
up vote 3 down vote accepted

Given two permutation groups H on n points and K on m points, there is a permutation group called H × K that acts on n + m points. The group is abstractly the direct product of the two groups, and the action is very simple: an element like $(h,k)$ acts on the first n points exactly like h did on its n points, and on the last m points exactly like k did on its m points.

For instance the Sylow 3-subgroup of Sym(6) is $\langle (1,2,3) \rangle \times \langle (4,5,6) \rangle$.

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Thanks Jack. So based on this new action, I think I can view the product as isomorphic to a subgroup of $S_{a_1p+a_2p^2+\cdots+a_kp^k}$, but where did the $a_0$ term go? Maybe I applied it incorrectly. – Nastassja Jun 8 '12 at 21:14
Exactly. As for $a_0$: the Sylow 17-subgroup of Sym(6) is ... the identity. The $a_0$ is the number of points that don't get moved at all by a Sylow $p$-subgroup. – Jack Schmidt Jun 8 '12 at 21:18
Ok, so we can always just include $a_0$ other points unaffected by the Sylow $p$-subgroup to view it as isomorphic to a subgroup of $S_{a_0+a_1p+\cdots+a_kp^k}=S_n$? – Nastassja Jun 8 '12 at 21:21
Yes exactly. Similar to how Qiaochu Yuan said, there is always an embedding of $S_m$ into $S_{m+a_0}$, just leave the last $a_0$ points alone. – Jack Schmidt Jun 8 '12 at 21:25

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