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Find the order of a group $Aut(\mathbb{Z}_3 \times \mathbb{Z}_9)$. Is it abelian? ($Aut(G)$ is a group of all automorphisms on $G$)

What i did:

Firstly, $\mathbb{Z}_3 \times \mathbb{Z}_9$ = $\langle (1,\: 0), (0,\: 1) \rangle$. If $\varphi \in Aut(\mathbb{Z}_3 \times \mathbb{Z}_9) $ than $ord\: \varphi( (1,\: 0) ) = 3$ and $ord\: \varphi( (0,\: 1) ) = 9$. There are $8$ elements in $\mathbb{Z}_3 \times \mathbb{Z}_9$ whose $ord$ is $3$ and $18$ elements whose $ord$ is $9$ (easy comb and $27 - 9 = 18$, $9 - 1 = 8$). So, $|Aut(\mathbb{Z}_3 \times \mathbb{Z}_9)| = 18 \cdot 8 = 2^{4} 3^{2}$

And i actually have no idead how to answer the second question. Any hints or sollutions, pls?

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    I'm not sure about your calculation of $|Aut(\mathbb{Z}_3\times\mathbb{Z}_9)|$. You can't simply send $(1,0)$ to any element of order $3$ and $(0,1)$ to any element of order $9$. e.g. $\varphi((1,0))= (0,3)$, $\varphi((0,1))= (0,1)$, though a homomorphism, is not an automorphism. – 2017-01-18

1 Answers 1

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An automorphism must preserve the direct product structure, so once you map $(0,1)$ to some element $y$ of order $9$, you must map $(1,0)$ to some element $x$ of order $3$ other than the two elements of order $3$ in the cyclic subgroup generated by $y$. Thus, while you have $18$ choices for $y$, once $y$ is selected, you only have $6$ choices for $x$, not $8$. It's clear that for any such choice of $x$ and $y$, the direct product structure is preserved, hence $|\text{Aut}(Z_3{\times}Z_9)| = 6\times18 = 108$.

For the question of whether $\text{Aut}(Z_3{\times}Z_9)$ is abelian, my guess is "no" by the guiding principle that functions tend not to commute unless forced. To test it, try some compositions, and see if the result is the same when you change the order of composition.