Sylow $p$-subgroup of $\operatorname{Sym}(p^2)$

Find generators for a Sylow $p$-subgroup of $\operatorname{Sym}(p^2)$: $p$ is a prime. Show that this is a non-abelian group of order $p^{p+1}$. (Abstract Algebra: Dummit & Foote, Sylow's theorem, Ex. 46)

Refer to: Sylow $p$ subgroups of $S_{2p}$ and $S_{p^2}$ - Dummit foote - $4.5.45; 4.5.46$, where only odd primes $p$ were concerned. I actually began by investigating the case when $p=2$. Sylow $2$-subgroups of $\operatorname{Sym}(4)$ are non-abelian groups of order $8$, all of which are $\cong \operatorname{Dih}(4)$:e.g. $\langle (1234), (13)\rangle$.

For $p=3$, Sylow 3-subgroups of $\operatorname{Sym}(9)$ have order $81$. I suspect that $\langle(123456789), (147)\rangle$ is one of them (just plain guess), but I'm not able to even compute the order of this subgroup. How do I know if I'm correct or not (without having to refer to the subgroup $\langle(123)(456)(789), (147)(258)(369)\rangle$ stated in the next paragraph)?

I found, from the link above, that another option for a Sylow $3$-subgroup of $\operatorname{Sym}(9)$ is indeed $\langle(123)(456)(789), (147)(258)(369)\rangle$ (and I found that $(123456789), (147)$ generate the generators of this subgroup). This is referred to as a 'wreath product of $\mathbb Z_3$ and $\mathbb Z_3$' and I found on Groupprops that $\operatorname{Dih}(4)$ is also a 'wreath product of $\mathbb Z_2$ and $\mathbb Z_2$'! In general, is there a set of generators and relations of these 'wreath products' for me to use, so that I can check my work?

You can check $G:=\langle(1,2,\ldots,9),(1,4,7)\rangle$ has order $81$ as follows:

Set $x=(1,2,\ldots,9)$. You have $(1,4,7),(2,5,8)=(1,4,7)^x,(3,6,9)=(1,4,7)^{x^2}\in G$. Therefore $N:=\langle(1,4,7),(2,5,8),(3,6,9)\rangle\le G$. You can check $N$ is in fact normal in $G$. In particular $G=N\langle x\rangle$, so $|G|=|N||\langle x\rangle|/|N\cap\langle x\rangle|$. $N\cong C_3^3$ and $x^3\in N$ so $|G|=27\cdot 9/3=81$.

The Wikipedia page for the wreath product shows you how to construct the wreath product in general. For $S_{p^2}$ take the base group $N=\langle(1,2,\ldots,p),(p+1,p+2,\ldots,2p),\ldots,(p(p-1)+1,p(p-1)+2,\ldots,p^2)\rangle$, then the group $H=\langle (1,p+1,\ldots,p(p-1)+1)(2,p+2,\ldots,p(p-1)+2)\cdots(p,2p,\ldots,p^2)\rangle$. $G=NH$ is then your wreath product.

Bonus: If you do get your head around the wreath product you may be interested to know that the Sylow subgroups of $S_{p^n}$ are isomorphic to the repeated wreath product $C_p\wr C_p\wr\cdots\wr C_p$ (read $C_p$ $n$ times).

• $(258)=(147)^x$ and $(369)=(147)^{x^2}$ seem weird to me. Could you explain a little bit more? – user441558 Jun 30 '17 at 7:10
• oh, of course. This actually depends a bit on convention - I mean the conjugate $(1,4,7)^x=x^{-1}(1,4,7)x$. This assume $S_{p^2}$ is acting on $\{1,\ldots,p^2\}$ on the right. If you have it acting on the left, then replace $(1,4,7)^x$ with $x(1,4,7)x^{-1}$. – Robert Chamberlain Jun 30 '17 at 7:18
• How do you know that $G=N\langle x\rangle$ in the first place? D: – user441558 Jun 30 '17 at 7:39
• Of the two generators you gave, one appears in $N$ and one appears in $\langle x\rangle$. Therefore $G\le N\langle x\rangle$. My answer shows the opposite inclusion so $G=N\langle x\rangle$. – Robert Chamberlain Jun 30 '17 at 7:41
• the intuition for this proof is to take the normal closure of one of the generators (in this case obtaining $N$) then hope that $G$ breaks down as a product of subgroups somehow – Robert Chamberlain Jun 30 '17 at 7:44