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

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.

## Author: mathtuition88

https://mathtuition88.com/

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