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Consider $H = \mathbb{Z}_{30}$ and $G = \mathbb{Z}_{15}$ as additive abelian groups. Then how do I show that ${\rm Aut}(H) \cong {\rm Aut}(G)$?

By the Chinese remainder theorem, I know that $\mathbb{Z}_{30} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{15}$. Intuitively I know that the only automorphism of $\mathbb{Z}_{2}$ is the identity automorphism, so we can construct a bijective map between $\psi : {\rm Aut}(G) \rightarrow {\rm Aut}(H)$. By $\psi(\phi) = \phi^{*}$, where $\phi^{*}((a,b)) = (a, \phi(b))$, which is an isomorphism. (Here we consider $\mathbb{Z}_{30} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{15}$.)

Is the definition of this map sufficient to claim that ${\rm Aut}(G) \cong {\rm Aut}(H)$? Also, how does one in general find the automorphism group of $\bigoplus_{k=1}^{r} \mathbb{Z}_{n_{k}}$, where the $n_{k}$'s are not necessarily coprime.

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Hint: $${\rm Aut}(\Bbb Z_n)\cong U(n),$$

where $U(n)$ is the group of units modulo $n$.

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For your first question, we have $\Bbb Z_{30}^×\cong(\Bbb Z_2\times\Bbb Z_{15})^×\cong\Bbb Z_2^×\times\Bbb Z_{15}^×\cong\Bbb Z_{15}^×$.

And $\operatorname {Aut}(\Bbb Z_n)\cong\Bbb Z_n^×$.

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