## Basic properties of wreath products

Let $K$ and $L$ be groups, let $n$ be a positive integer, let $\rho : K \rightarrow S_n$ be a homomorphism, and let $H = L^n$. In this previous exercise, we defined an injective homomorphism $\psi : S_n \rightarrow \mathsf{Aut}(H)$ by permuting the factors. Now $\psi \circ \rho$ is a homomorphism $K \rightarrow \mathsf{Aut}(H)$. The wreath product of $L$ by $K$ via $\rho$ is defined $L \wr_\rho K = H \rtimes_{\psi \circ \rho} K$. Note that the wreath product depends on the choice of permutation representation; if none is given, $\rho$ is assumed to be the left regular representation.

1. Assume $L$ and $K$ are finite and $\rho$ is the left regular representation of $K$. Find $|L \wr_\rho K|$ in terms of $|K|$ and $|L|$.
2. Let $p$ be a prime, let $K = L = Z_p$, and let $\rho$ be the left regular representation of $K$. Prove that $Z_p \wr_\rho Z_p$ is a nonabelian group of order $p^{p+1}$ and is isomorphic to a Sylow $p$-subgroup of $S_{p^2}$. [The $p$ copies of $Z_p$ whose direct product makes up $H$ may be represented by $p$ disjoint $p$-cycles; these are cyclically permuted by $K$.]

1. We have $L \wr_\rho K = L^{|K|} \rtimes K$. Thus $|L \wr_\rho K| = |L|^{|K|} \cdot |K|$.
2. Since the left regular representation is nontrivial, and $\psi$ is nontrivial, $Z_p \wr_\rho Z_p$ is a nonabelian group. By part (1), this group has order $p^p \cdot p = p^{p+1}$.

Now $H = Z_p^p$ is a finite abelian group. Let $\sigma_i$, $1 \leq i \leq p$, be $p$ disjoint $p$-cycles in $S_{p^2}$. By this previous exercise, there is a unique group homomorphism $\varphi : H \rightarrow S_{p^2}$ such that $\varphi(e_i) = \sigma_i$, where $e_i$ is the $i$th “standard basis” vector. Moreover, $(\psi \circ \rho)(k)(e_i) = e_{i+k}$. By our solution to this previous exercise, there exists a permutation $\tau \in S_{p^2}$ such that $\tau \sigma_i \tau^{-1} = \sigma_{i+1}$, where indices are taken mod $p$. Define $\theta : Z_p \wr_\rho Z_p \rightarrow S_{p^2}$ by $\theta(h,k) = \varphi(h)\tau^{-k}$.

I claim that $\theta$ is a homomorphism. To see this, write $h_1 = \prod e_i^{a_i}$ and $h_2 = \prod e_i^{b_i}$. Then $\theta((h_1,k_1)(h_2,k_2)) = \theta(((\prod e_i^{a_i})(\psi \circ \rho)(k_1)(\prod e_i^{b_i}), k_1k_2))$ $= \theta(((\prod e_i^{a_i})(\prod e_{i+k_1}^{b_i}), k_1k_2))$ $= \theta((\prod e_i^{a_i + b_i - k_1}), k_1 k_2)$ $= \prod \sigma_i^{a_i + b_i - k_1} \tau^{-k_1-k_2}$ $= (\prod \sigma_i^{a_i})\tau^{-k_1} (\prod \sigma_i^{b_i}) \tau^{-k_2}$ $= \theta(h_1,k_1) \theta(h_2,k_2)$.

Moreover, I claim that $\theta$ is injective; in this previous exercise we saw that every element of $\mathsf{im}\ \theta$ can be written uniquely as a product $(\prod \sigma_i^{a_i}) \tau^k$. Thus $\theta$ is injective. Counting elements, we see that $\mathsf{im}\ \theta$ is a Sylow $p$-subgroup of $S_{p^2}$.