How could one write $Aut(C_{30})$ as a product of cyclic groups?

Question

I am somewhat familiar with determining the product form of an arbitrary cyclic group but not of the automorphism of a cyclic group.

Thanks in advance.

Answer 1

First of all let us choose a generator $g\in C_{30}$, and view it as $\mathbb{Z}/{30\mathbb{Z}}$.

Any automorphism is determined by the image of the class $[1]$, so the endomorphisms are given by $$ \text{End}(\mathbb{Z}/30\mathbb{Z}) = \{[1]\mapsto[n]\ \mid n\in \mathbb{Z}\}. $$

The multiplication structure of composing endomorphisms comes down to multiplying the integers $n$, so the automorphisms are exactly the endomorphisms such that the class $[n]$ is invertible mod 30.

With this knowledge we find $$ \text{Aut}(\mathbb{Z}/30\mathbb{Z}) \cong (\mathbb{Z}/30\mathbb{Z})^*. $$

By the chinese remainder theorem, $$ (\mathbb{Z}/30\mathbb{Z})^* \cong (\mathbb{Z}/5\mathbb{Z})^* \times (\mathbb{Z}/2\mathbb{Z})^* \times (\mathbb{Z}/3\mathbb{Z})^* \cong (\mathbb{Z}/4\mathbb{Z})^* \times (\mathbb{Z}/2\mathbb{Z})^*. $$

EDIT: replaced automorphism with endomorphism.

Answer 2

$$\text{Aut}(C_{30})\simeq \mathbb{Z}/(30\mathbb{Z})^* \stackrel{\text{CRT}}{\simeq}\mathbb{Z}/(2\mathbb{Z})^*\times \mathbb{Z}/(3\mathbb{Z})^*\times \mathbb{Z}/(5\mathbb{Z})^*\simeq\color{red}{C_2\times C_4}.$$

Answer 3

First you might want to know the order is $\phi(30)=8$ as per this post. Essentially you have to map a generator to a generator in order for the induced homomorphism to be an automorphism and there's $\phi(30)$ choices.

Then suppose $C_{30}=\langle a\rangle$, the maps are $\phi_i:a\mapsto a^i$, then $\phi_i \circ \phi_j=\phi_j \circ \phi_i$, just by checking what happens to $a$, so the group is abelian and you have only three choices: $\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$, $\Bbb Z_2\times \Bbb Z_4$ and $\Bbb Z_8$.

You can kind of already see that $\operatorname{Aut}(C_{30})$ will be isomorphic to the units of $\Bbb Z_{30}$ i.e. $(\Bbb Z_{30})^*$ just by the way that $\phi_i\circ\phi_j(a)=(a^{j})^i=a^{i\cdot j}$.

Anyway the $8$ numbers co-prime to $30$ are $1,7,11,13,17,19,23,29$, and it's easy to check that $\phi_7$ has order $4$, since modulo $30$ multiplication by seven follows $7\to 19\to 13\to 1$ which rules out $\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$. It's also easy to check that $\phi_{11}$ has order $2$, and $\phi_{17}$ has order $4$, since $17\to 19\to 23\to 1$. So counting the order $4$ elements there are $4$, namely $\phi_{7},\phi_{13},\phi_{17},\phi_{23}$, whereas $\Bbb{Z}_8$ only has $2$, namely $2,6$.

So it has to be $\Bbb Z_2\times \Bbb Z_4$.

This is a bit of a lower level approach than the other answers depending on how one knows all of the abelian groups of order $8$.