Aut(G)=Aut(H)xAut(K), where H, K are characteristic subgroups of G with trivial intersection

Posted on October 17, 2016 by

Let G=HK, where H, K are characteristic subgroups of G with trivial intersection, i.e. H\cap K=\{e\}. Then, Aut(G)=Aut(H)\times Aut(K).

Proof:

Now suppose G=HK, where H and K are characteristic subgroups of G with H\cap K=\{e\}. Define \Psi:\text{Aut}(G)\to\text{Aut}(H)\times\text{Aut}(K) by \displaystyle \Psi(\sigma)=(\sigma|_H, \sigma|_K).

\sigma|_H:H\to H is a homomorphism, and bijective since \sigma|_H(H)=H. Thus \sigma|_H\in\text{Aut}(H) and similarly, \sigma|_K\in\text{Aut}(K) so that \Psi is well-defined.

Note that \displaystyle \Psi(\sigma_1\sigma_2)=((\sigma_1\sigma_2)|_H, (\sigma_1\sigma_2)|_K)=(\sigma_1|_H,\sigma_1|_K)(\sigma_2|_H,\sigma_2|_K)=\Psi(\sigma_1)\Psi(\sigma_2) so \Psi is a homomorphism.

Suppose \sigma\in\ker\Psi. Then \Psi(\sigma)=(\sigma|_H,\sigma|_K)=(\text{id}_H,\text{id}_K). Then for hk\in G, \sigma(hk)=\sigma(h)\sigma(k)=hk so that \sigma=\text{id}_G. Thus \Psi is injective.

For any (\phi, \psi)\in\text{Aut}(H)\times\text{Aut}(K), define \sigma(hk)=\phi(h)\psi(k). Then
\begin{aligned} \sigma(h_1k_1h_2k_2)&=\sigma(h_1h_2k_1k_2)\\ &\text{(}H, K\ \text{normal and}\ H\cap K=\{e\}\ \text{implies elements of}\ H, K\ \text{commute)}\\ &=\phi(h_1h_2)\psi(k_1k_2)\\ &=\phi(h_1)\phi(h_2)\psi(k_1)\psi(k_2)\\ &=\phi(h_1)\psi(k_1)\phi(h_2)\psi(k_2)\\ &=\sigma(h_1k_1)\sigma(h_2k_2). \end{aligned}
So \sigma is a homomorphism.

If hk\in\ker\sigma, then \phi(h)\psi(k)=e, so that \phi(h)=(\psi(k))^{-1}. Then since H\cap K=\{e\}, so \phi(h)=\psi(k)=e, so that h=k=e. Thus \ker\sigma=\{e\} and \sigma is injective.

Any h\in H can be written as \phi(h') since \phi is bijective. Similarly, any k\in K can be written as \psi(k'). Then \sigma(h'k')=\phi(h')\psi(k')=hk so \sigma is surjective.

Thus \sigma\in\text{Aut}(G). Note that \sigma|_H=\phi since \sigma|_H(h)=\sigma(h\cdot 1)=\phi(h)\psi(1)=\phi(h). Similarly, \sigma|_K=\psi. So \Psi(\sigma)=(\phi,\psi) and \Psi is surjective.

Hence \Psi is an isomorphism.

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