Some basic properties of holomorphs

Let H be a group of order n, let K = \mathsf{Aut}(H), and form G = \mathsf{Hol}(H) = H \rtimes_1 K. Let G act by left multiplication on the left cosets of K in G and let \pi : G \rightarrow S_{G/K} be the induced permutation representation.

  1. Prove that the elements of H are coset representatives for the left cosets of K in G and that with this choice of representatives, \pi|_H is the left regular representation of H.
  2. Prove that \mathsf{im}\ \pi is the normalizer in S_{G/K} of \pi[H]. Deduce that under the left regular representation of a finite group H of order n, the noramlizer in S_n of the image of H is isomorphic to \mathsf{Hol}(H). [Hint: Show that |G| = |N_{S_n}(\pi[H])|] using §5.5 #1 and §5.5 #2.
  3. Deduce that the normalizer of the group generated by an n-cycle in S_n is isomorphic to \mathsf{Hol}(Z_n) and has order n \varphi(n), where \varphi denotes the Euler totient function.

  1. Recall that (h,k) = (h,1)(1,k). It suffices to show that if h_1,h_2 \in H such that h_1K = h_2K, then h_1 = h_2. To that end, if (h_1,1) = (h_2,k), then h_1 = h_2. There are n left cosets of K, and H maps to these injectively. Thus H is a set of coset representatives of G/K. Define \theta : S_{G/K} \rightarrow S_H by \theta(hK) = h, and let \rho : H \rightarrow S_H be the left regular representation of H. Then \rho(h)(a) = ha = (ha)K h(aK) = \pi(h)(aK) = (\pi(h) \circ \theta)(a). Hence \rho(h) = \pi(h) \circ \theta; thus \pi[H] is the left regular representation of H up to an isomorphism.
  2. First, let \pi(g) \in \pi[G]. Now \pi(g)\pi[H]\pi(g)^{-1} = \pi[gHg^{-1}] = \pi[H]. Thus \pi(g) \in N_{S_{G/K}}(\pi[H]).

    Now we prove a lemma.

    Lemma [Adapted from Wikipedia]: If \sigma \in N_{S_{G/K}}(\pi[H]) and \sigma(1) = 1, then \sigma \in \mathsf{Aut}(H). Proof: Since \sigma normalizes \pi[H], for every h \in H there exists k \in H such that \sigma \circ \pi(h) = \pi(k) \circ \sigma. Applying this mapping to 1, we see that \sigma(h) = k. Thus, if \sigma(1) = 1, then for all h \in H we have \sigma \circ \pi(h) = \pi(\sigma(h)) \circ \sigma. Now for all h,k \in H, we have \sigma \circ \pi(h) \circ \pi(k) = \pi(\sigma(h)) \circ \sigma \circ \pi(k) = \pi(\sigma(h)) \circ \pi(\sigma(k)) \circ \sigma = \pi(\sigma(h) \sigma(k)) \circ \sigma, which is equal (since \pi is a homomorphism) to \sigma \circ \pi(hk) = \pi(\sigma(h,k)) \circ \sigma. Since \sigma is invertible and \pi is injective, we have \sigma(hk) = \sigma(h) \sigma(k); hence \sigma is a homomorphism H \rightarrow H. Since \sigma is also bijective, \sigma \in \mathsf{Aut}(H). \square

    (\supseteq) Now let \sigma \in N_{S_{G/K}}(\pi[H]). By the argument above, \pi(\sigma(1)^{-1}) also normalizes \pi[H]. Since the normalizer is a subgroup, we have \pi(\sigma^{-1}(1)) \circ \sigma \in N_{S_{G/K}}(\pi[H]). Moreover, (\pi(\sigma^{-1}(1)) \circ \sigma)(1) = \pi(\sigma(1)^{-1})(\sigma(1)) = 1; by the lemma, \beta = \pi(\sigma^{-1}(1)) \circ \sigma is an automorphism of H. Finally, note that \pi(\beta)(h) = \beta(h), so that \beta = \pi(\beta). Thus we have \sigma = \pi(\sigma(1)^{-1})^{-1} \circ \beta = \pi(\sigma(1)) \circ \pi(\beta) = \pi(\sigma(1),\beta). Hence \sigma \in \pi[G].

    Since \pi is injective, we conclude that \mathsf{Hol}(H) \cong N_{S_H}(\mathsf{im}\ \rho).

  3. Let \tau be an n-cycle in S_n. Up to a renaming, \langle \tau \rangle is the image of Z_n in S_{Z_n} via the permutation representation \pi : \mathsf{Hol}(Z_n) \rightarrow S_n used above. Now N_{S_n}(\langle \tau \rangle) \cong \mathsf{Hol}(Z_n), and \mathsf{Hol}(Z_n) = Z_n \rtimes \mathsf{Aut}(Z_n). We know that |\mathsf{Aut}(Z_n)| = \varphi(n), where \varphi denotes the Euler totient. Thus |N_{S_n}(\langle \tau \rangle)| = n\varphi(n).
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