This is Exercise 5.3.2 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE. The result is mentioned in the following posts though:
- This answer to $p$-Sylow subgroup of $S_{p^3}$
- This answer to On the number of Sylow subgroups in Symmetric Group
- On $p$-Sylows for $S_{2p}$ … $S_{(p-1)p}$
- When is normalizer of wreath product the wreath product of normalizers?
Here are some previous questions of mine about Wreath products:
- Showing $Z(H\wr K)$, for abelian $H\neq 1$ and arbitrary $K$, is the diagonal subgroup of the base group.
- Showing $Z(H\wr K)=1$, for abelian $H\neq 1$ and arbitrary, infinite $K$.
- Showing that the restricted wreath product $\Bbb Z\wr\Bbb Z$ is finitely generated.
- The Frattini subgroup of the standard Wreath product of two quasicyclic groups is the group itself.
- There exist infinite solvable $p$-groups with trivial centre. (Use a hint.)
The Details:
(This can be skipped.)
On page 32 and 33 of the book,
Let $H$ and $K$ be permutation groups on sets $X$ and $Y$ respectively. [. . .]
If $\gamma\in H$, $y\in Y$, and $\kappa\in K$, define
$$\gamma(y):\begin{cases} (x,y) \mapsto (x\gamma, y) & \\ (x,y_1) \mapsto (x,y_1) & y_1\neq y,\end{cases}$$
and$$\kappa^\ast: (x,y)\mapsto (x,y\kappa).$$
[. . .] The functions $\gamma\mapsto \gamma(y)$, with $y$ fixed, and $\kappa\mapsto \kappa^\ast$ are monomorphisms from $H$ and $K$ to ${\rm Sym}(X\times Y)$: let their images be $H(y)$ and $K^\ast$, respectively. Then the wreath product of $H$ and $K$ is the permutation group on $X\times Y$ [. . .]
$$H\wr K=\langle H(y), K^\ast\mid y\in Y\rangle.$$
On page 41 of the book,
If $H$ and $K$ are arbitrary groups, we can think of them as permutation groups on their underlying sets via the right regular representation and form their wreath product $W=H\wr K$: this is called the standard wreath product.
The Question:
A group of order $p^n$ is isomorphic with a subgroup of the standard wreath product $$\underbrace{\Bbb Z_p\wr\dots\wr\Bbb Z_p.}_{n\text{ factors}}$$
Here $p$ is prime.
Thoughts:
Since semidirect products are associative and each wreath product can be viewed as a semidirect product, the lack of parentheses in question makes sense.
Fix a prime $p$.
My first idea was to use induction on $n$, since for the base case $n=1$, we have that the group in question is $\Bbb Z_p$.
I didn't get anywhere here though.
Perhaps we could modify a proof of Cayley's theorem to do the exercise. Its statement:
Each group is isomorphic to a subgroup of a symmetric group.
Another useful theorem might be Sylow's First Theorem:
For every prime factor $p$ with multiplicity $n$ of the order of a finite group $G$, there exists a Sylow $p$-subgroup of $G$, of order $p^n$.
A Sylow $p$-subgroup of a group $G$ is subgroup of $G$ of order $p^k$ for $k$ maximal such that $p^k\mid\lvert G\rvert$.
Please help 🙂
Best Answer
By Cayley's Theorem, we can embed $G$ into the symmetric group $S_{|G|} = S_{p^n}$. So by Sylow's Theorem, $G$ is isomorphic to a subgroup of a Sylow $p$-subgroup $P$ of $S_{p^n}$.
(I prefer $C_p$ to ${\mathbb Z}_p$ for a cyclic group of order $p$.) It is a standard result that $P \cong C_p \wr C_p \wr \cdots \wr C_p$ (with $n$ wreath factors.) Let's write that as $C_p^{\wr^n}$.
That is not hard to prove by induction. Assuming that $S_{p^{n-1}}$ has Sylow $p$-subgroup $Q \cong C_p^{\wr^{n-1}}$, by partitioning the set $\{1,2,\ldots,p^n\}$ into $p$ sets of size $p^{n-1}$, we see that $S_{p^n}$ contains the permutation wreath product $Q \wr C_p \cong C_p^{\wr^n}$ and by computing its order you can check that it is a Sylow $p$-subgroup.