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Can you help me with the following problem?

Let $H,K$ be normal subgroups of group $G$, if $\gcd\left(|H|,|K|\right)=1$ then $\text{Aut}(H \times K)=\text{Aut}(H)\times \text{Aut}(K)$.

Is it true that $\text{Aut}(H)\times\text{Aut}(K)$ is contained in $\text{Aut}(H \times K)$ regardless of coprimeness? Why?

How to use coprimeness to show the other containing?

Arbuja
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C.Tsioutis
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For the first part, yes, we can identify $\mathrm{Aut}(H)\times\mathrm{Aut}(K)$ with a subgroup of $\mathrm{Aut}(H\times K)$.

This is because any $f\in\mathrm{Aut}(H)$, $g\in\mathrm{Aut}(K)$ induce automorphisms $f',g'\in\mathrm{Aut}(H\times K)$ by $f'(h,k)=(f(h),k)$ and $g'(h,k)=(h,g(k))$. Clearly $f'$ and $g'$ always commute, so $\mathrm{Aut}(H)\times\mathrm{Aut}(K)\subseteq\mathrm{Aut}(H\times K)$ via $(f,g)\mapsto f'g'$.

For the second part, the opposite inclusion, consider $\rho\in\mathrm{Aut}(H\times K)$. For $h\in H$, $\rho(h)$ must have the same order as $H$. Since $H$ and $K$ have coprime order, the only elements with the same order as $h$ must be in $H$, so $\rho(H)=H$. In particular $\rho$ restricts to an automorphism $f\in\mathrm{Aut}(H)$. Similarly, $\rho$ restricts to some $g\in\mathrm{Aut}(K)$. Noticing then that $\rho=f'g'$, we obtain $\rho\in\mathrm{Aut}(H)\times\mathrm{Aut}(K)$.

Robert Chamberlain
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  • Doesn't this only prove that these automorphism groups are equal as sets? Why does this prove that they are isomorphic as groups? –  Feb 21 '19 at 00:44
  • the first part $(f,g)\mapsto f'g'$ defines an injective homomorphism, the second part shows that the homomorphism is surjective – Robert Chamberlain Feb 21 '19 at 07:59