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 ith “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}.

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