If the direct factors are characteristic in a direct product, then the automorphism group of the product is the product of automorphism groups

Prove that if H and K are groups such that H \times 1 and 1 \times K are characteristic in H \times K, then \mathsf{Aut}(H \times K) \cong \mathsf{Aut}(H) \times \mathsf{Aut}(K). Deduce that if G is an abelian group of finite order then \mathsf{Aut}(G) is isomorphic to the direct product of the automorphism groups of its Sylow subgroups.

We define a mapping \Phi : \mathsf{Aut}(H) \times \mathsf{Aut}(K) \rightarrow \mathsf{Aut}(H \times K) as follows: \Phi(\varphi, \psi) = \varphi \times \psi. We claim that \Phi is an isomorphism.

  1. (\Phi is a homomorphism) Suppose all mappings are in the appropriate sets. Then \Phi((\varphi_1,\psi_1)(\varphi_2,\psi_2)) = \Phi(\varphi_1 \circ \varphi_2, \psi_1 \circ \psi_2) = (\varphi_1 \circ \varphi_2) \times (\psi_1, \circ \psi_2) = (\varphi_1 \times \psi_1) \circ (\varphi_2 \times \psi_2) = \Phi(\varphi_1, \psi_2) \circ \Phi(\varphi_2, \psi_2).
  2. (\Phi is injective) Suppose \Phi(\varphi_1,\psi_1) = \Phi(\varphi_2,\psi_2). Then for all h \in H and k \in K, we have (\varphi_1(h), \psi_1(k)) = (\varphi_1 \times \psi_1)(h,k) = (\varphi_2 \times \psi_2)(h,k) = (\varphi_2(h), \psi_2(k)). Since pairs are equal precisely when their corresponding entries are equal, we have \varphi_1(h) = \varphi_2(h) and \psi_1(k) = \psi_2(k) for all h \in H and k \in K; thus \varphi_1 = \varphi_2 and \psi_1 = \psi_2, and we have (\varphi_1,\psi_1) = (\varphi_2,\psi_2).
  3. (\Phi is surjective) Let \chi \in \mathsf{Aut}(H \times K). Let h \in H; since H \times 1 \leq H \times K is characteristic, \chi(h,1) \in H \times 1. Define a mapping \chi_H : H \rightarrow H by \chi(h,1) = (\chi_H(h), 1). We claim that \chi_H is an automorphism of H.
    1. (\chi_H is a homomorphism) Note that (\chi_H(h_1h_2), 1) = \chi(h_1h_2,1) = \chi(h_1,1) \chi(h_2,1) = (\chi_H(h_1),1) (\chi_H(h_2), 1) = (\chi_H(h_1) \chi_H(h_2), 1). Comparing entries, we have \chi_H(h_1h_2) = \chi_H(h_1) \chi_H(h_2) for all h_1,h_2 \in H.
    2. (\chi_H is injective) Suppose \chi_H(h_1) = \chi_H(h_2). Then \chi(h_1,1) = (\chi_H(h_1),1) = (\chi_H(h_2),1) = \chi(h_2,1). Since \chi is injective, h_1 = h_2.
    3. (\chi_H is surjective) Let h_1 \in H. Since \chi[H \times 1] = H \times 1, there exists h_2 \in H such that \chi(h_2,1) = (h_1,1). Thus h_1 = \chi_H(h_2).

    Likewise, we can define a mapping \chi_K : K \rightarrow K which is also an isomorphism.

    Finally, note that \chi(h,k) = \chi((h,1)(1,k)) \chi(h,1)\chi(1,k) = (\chi_H(h),1)(1,\chi_K(k)) = (\chi_H(h), \chi_K(k)) = (\chi_H \times \chi_K)(h,k) Thus \chi = \chi_H \times \chi_K, and we have \chi = \Phi(\chi_H, \chi_K).

Thus \mathsf{Aut}(H \times K) \cong \mathsf{Aut}(H) \times \mathsf{Aut}(K).

As a lemma, we can extend this result by induction to finite direct products in which each factor is characteristic.

Lemma: Let G_i be groups such that G_{\{i\}} (c.f. §5.1 #2) is characteristic in G = \times_{i=1}^n G_i for each i. Then \mathsf{Aut}(\times_{i=1}^n G_i) \cong \times_{i=1}^n \mathsf{Aut}(G_i). Proof: We proceed by induction on n. The conclusion is trivial for n=1, and we proved the n=2 case above. Suppose the conclusion holds for some n \geq 2. Now \times_{i=1}^{n+1} G_i \cong (\times{i=1}^n G_i) \times G_{n+1}. Suppose \varphi is an automorphism of G; then \varphi[\times_{i=1}^n G_i] = \varphi[\prod_{i=1}^n G_{\{i\}}] = \prod_{i=1}^n \varphi[G_{\{i\}}] = \prod_{i=1}^n G_{\{i\}} = \times_{i=1}^n G_i, so that \times_{i=1}^n G_i is characteristic. By the induction hypothesis and part (a) of this problem, we have \mathsf{Aut}(\times_{i=1}^{n+1} G_i) \cong \times_{i=1}^{n+1} \mathsf{Aut}(G_i). \square

Now let A be a finite abelian group. Note that every Sylow subgroup P is normal, hence the unique subgroup of order |P|. Thus each Sylow subgroup is characteristic in A. In §5.4 #10, we saw that a finite abelian group is (isomorphic to) the direct product of its Sylow subgroups; the final conclusion then follows from the lemma.

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