Hint Request For: Finding Order $3$ Automorphisms of $Z_{91}$ Using Group Actions?

Question

I have been stuck on the following problem for a while:

Find all automorphisms of $Z_{91}$ of order $3$. [Hint: How can $Z_3$ act non-trivially on $Z_{91}$?]

Does $Z_{91}$ have any automorphisms of order $5$?

Could you please give me a hint? I am interested in any other (group-theoretic) way of doing the problem, but I am particularly interested in following the hint.

My Work: Let $G = Z_{91}$. I know that $\text{Aut}(G) \cong (\mathbb{Z}/91\mathbb{Z})^{\times}$ which has $\varphi(91)=72$ elements, so I can already tell that $Z_{91}$ has no automorphism of order $5$.

Since $\text{Aut}(G)$ is abelian, the set of all elements of $\text{Aut}(G)$ which satisfy $x^3=1$ form a subgroup.

Edit (Response to Jamie):

The only thing I could think to do using Sylow Theorems is to shot that the automorphism group has either $1$ or $4$ subgroups of order $9$, but I didn't really see where to go from there.

Since the question asks to find the automorphisms of order $3$ (not just the number of them), the only thing I could think of is to trace elements of order $3$ in $Z_6 \times Z_{12}$ back to elements of $\text{Aut}(Z_{91})$.

Let $Z_6 = \langle x \rangle$ and $Z_{12} = \langle y \rangle$. The only elements of order $3$ in $Z_6 \times Z_{12}$ are

$$ \begin{bmatrix} & (1, y^4) & (1, y^8)\\ (x^2, 1) & (x^2, y^4) & (x^2, y^8\\ (x^4, 1) & (x^4, y^4) & (x^4, y^8 \end{bmatrix} $$

Now, I found that $4$ generates $(\mathbb{Z}/7 \mathbb{Z})^{\times}$, and $2$ generates $(\mathbb{Z}/13 \mathbb{Z})^{\times}$ so for example, $(x^2, y^4)$ corresponds to the element $(4^2, 2^4) = (2, 3)$ in $(\mathbb{Z}/7 \mathbb{Z})^{\times} \times (\mathbb{Z}/13 \mathbb{Z})^{\times}$.

However, I'm don't know how to find the an explicit isomorphism between $(\mathbb{Z}/7 \mathbb{Z})^{\times} \times (\mathbb{Z}/13 \mathbb{Z})^{\times}$ and $(\mathbb{Z}/91 \mathbb{Z})^{\times}.$ If I have this, then I know how to make an explicit isomorphism from $(\mathbb{Z}/91 \mathbb{Z})^{\times} \to \text{Aut}(G).$ Could you please help me with that, or let me know if there is a different way using Sylow Theorems?

Answer 1

As already noted, $\operatorname {Aut}\Bbb Z_{91}\cong\Bbb Z_{91}^×\cong \Bbb Z_6×\Bbb Z_{12}$.

So the order $3$ automorphisms correspond to the elements $ (2,0),(2,4), (4,0), (4,4),(4,8), (2,8), (0,8)$ and $(0,4)$ of $\Bbb Z_6×\Bbb Z_{12}$.

Thus there are $8$.

To find the automorphisms explicitly, just find all $8$ order $3$ elements of $\Bbb Z_{91}^×$. Not sure how to do this, other than by hand, as i also don't have the explicit isomorphism between $\Bbb Z_{91}^× $ and $\Bbb Z_6\times \Bbb Z_{12}$ right here handy.

Since there are no elements of order five in $\Bbb Z_{91}^×$, there is no automorphism of order five.

Answer 2

Ignoring the hint, $(\mathbb{Z}/91\mathbb{Z})^{\times} \cong (\mathbb{Z}/13\mathbb{Z})^{\times} \times (\mathbb{Z}/7\mathbb{Z})^{\times}$ by the chinese reminder theorem. And $(\mathbb{Z}/13\mathbb{Z})^{\times} \times (\mathbb{Z}/7\mathbb{Z})^{\times} \cong Z_6 \times Z_{12}$ which has order 72. Now use Sylow's subgroup theorem noting $72=3^2*8$. Hopefully you can take it from there!