# Elements of order $3$ in $\text{Aut}\left(\mathbb{Z}/91\mathbb{Z}\right)$

It looks like someone has already been here, but the question I have goes farther.

To summarize my work, as well as the work in the above post, we know that $\text{Aut}\left(\mathbb{Z}/91\mathbb{Z}\right) \cong \mathcal{U}_{91} \cong \mathcal{U}_7 \times \mathcal{U}_{13}$ — by the Chinese Remainder Theorem. And $\mathcal{U}_7 \times \mathcal{U}_{13} \cong \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$.

For a while I was stuck where the other post happened to leave off: I found the eight elements of order $3$ in terms of $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$; great, but the question asks for automorphisms. I know that any $\varphi \in \text{Aut}\left(\mathbb{Z}/91\mathbb{Z}\right)$ must map a generator to a generator, so I assumed that $\varphi(1) = k$, where $(91, k) = 1$.

If $|\varphi| = 3$, then $\varphi^3(1) = k^3 = 1$ — i.e., $k^3 - 1 \equiv 0 \;(\text{mod } 91)$. Factoring $k^3 - 1$ as $(k-1)(k^2+k+1)$, I found that $k = 9$ and $k = 22$ gave solutions. So if $\pi(1) = 9$ and $\rho(1) = 22$, then $\pi, \rho \in \text{Aut}\left(\mathbb{Z}/91\mathbb{Z}\right)$, $|\pi| = |\rho| = 3$, and the elements of order $3$ are $\left< \pi, \rho \right> \backslash \{\text{id}\}$.

I was pretty happy with this, but then I started playing around with my answers, making sure they didn't lead to nonsense. The first problem I found was that, although $|26| = 7$ in $\mathbb{Z}/91\mathbb{Z}$, as defined above $\rho(26)=26\cdot\rho(1)=26\cdot 22 = 572 \equiv 4$, and $|4| = 91$. I tried redefining $\rho$ so that $\rho(1) = 29$ (as originally defined, this would have been $\rho^2(1)$, which of course also has order $3$), and that fixes my problem with $26$. But there are a few other numbers to check, and I'm not sure that checking them is a viable option.

So are $\pi$ and the new $\rho$ the automorphisms I want? If so, how can I be sure? — since my original $\rho$ broke down. If not, how do I find them, and, again, how do I know once I have?

I'll add that I suspect this may involve generators of $\mathcal{U}_7$ and of $\mathcal{U}_{13}$: $\mathcal{U}_7$ is generated by $3$, and $3^2 = 9$ — hence $\pi$ above. As for taking the fourth root of $29$ in $\mathcal{U}_{13}$ ... Even looking for a square root was taking too long.

You've essentially solved the problem by finding the elements of order $3$ in $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$, but now you need to chase back through the various isomorphisms to find the corresponding automorphisms.

You're right, to realize the isomorphism $(\mathbb{Z}/7\mathbb{Z})^\times \times (\mathbb{Z}/13\mathbb{Z})^\times \cong \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$, we need to find generators for $(\mathbb{Z}/7\mathbb{Z})^\times$ and $(\mathbb{Z}/13\mathbb{Z})^\times$. You found that $3$ generates $(\mathbb{Z}/7\mathbb{Z})^\times$ and since \begin{align*} 2^4 &\equiv 16 \equiv 3 \pmod{13}\\ 2^6 &\equiv 3 \cdot 4 \equiv 12 \pmod{13} \end{align*} then $2$ generates $(\mathbb{Z}/13\mathbb{Z})^\times$. This gives us the following isomorphism. \begin{align*} \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z} &\overset{\sim}{\to} (\mathbb{Z}/7\mathbb{Z})^\times \times (\mathbb{Z}/13\mathbb{Z})^\times\\ (a,b) &\mapsto (3^a, 2^b) \end{align*}

Next we need to find the isomorphism $(\mathbb{Z}/7\mathbb{Z})^\times \times (\mathbb{Z}/13\mathbb{Z})^\times \overset{\sim}{\longrightarrow} (\mathbb{Z}/91\mathbb{Z})^\times$. Since $1 = 14 - 13 = 2 \cdot 7 + (-1) 13$, then we have an isomorphism \begin{align*} (\mathbb{Z}/7\mathbb{Z})^\times \times (\mathbb{Z}/13\mathbb{Z})^\times &\overset{\sim}{\longrightarrow} (\mathbb{Z}/91\mathbb{Z})^\times\\ (r, s) &\longmapsto 14s - 13r \, . \end{align*} (See here for more on this isomorphism.)

Lastly, we have an isomorphism \begin{align*} (\mathbb{Z}/91\mathbb{Z})^\times &\overset{\sim}{\to} \text{Aut}(\mathbb{Z}/91\mathbb{Z})\\ k &\mapsto \varphi_k \end{align*} where $\varphi_k: \mathbb{Z}/91\mathbb{Z} \to \mathbb{Z}/91\mathbb{Z}$ is given by $\varphi_k(x) = kx$.

Taking the elements of order $3$ you found in $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$ and using these isomorphisms, you should be able to find the automorphisms of $\mathbb{Z}/91\mathbb{Z}$ of order $3$.

Note SpamIAm's method yields:

$(2,0) \mapsto \phi_{79}$ and $(0,4) \mapsto \phi_{29}$.

Explicitly calculating $\langle \phi_{79},\phi_{29}\rangle$, we find the subgroup of $\text{Aut}(\Bbb Z_{91})$ isomorphic to $\Bbb Z_3 \times \Bbb Z_3$ is:

$\{\phi_1,\phi_{79},\phi_{53},\phi_{29},\phi_{22},\phi_{16},\phi_9,\phi_{81},\phi_{74}\}$

So your $\phi_9$ and $\phi_{22}$ are indeed generators-your error was that $572 = 26$ (mod $91$), and clearly $26$ has order $\dfrac{91}{\gcd(26,91)} = 7$.

• Lol, yup, I did. Fixed. Jul 25 '15 at 7:16