Find elements of a group $\operatorname{Aut}(\mathbb{Z_{20}})$ automorphisms of cyclic group $\mathbb{Z_{20}}$

Question

Find elements of a group $\operatorname{Aut}(\mathbb{Z_{20}})$ automorphisms of cyclic group $\mathbb{Z_{20}}$. Is $\operatorname{Aut}(\mathbb Z_{20})$ cyclic group?

$\mathbb{Z_{20}} = \left\{0,1,...,19\right\}$

We know that an automorphism is an isomorphism from group G into the same group.

Let $\phi$ be random element of $\operatorname{Aut}(\mathbb{Z_{20}})$. Group $\mathbb{Z_{20}}$ is generated by $1$ and $\phi$ is a group homomorphism so $\phi$ is uniquely designated via image on element $1$.

Automorphism $\phi$ doesn't change group order so $\phi(1)$ must be of order $20$. So $|\operatorname{Aut}(\mathbb{Z_{20}})| = 20$.

These are my only observations but I don't know what to do next to get to specific elements

Please rethink the number of choices of $\phi(1)$. Many elements of $\Bbb{Z}_{20}$ have orders less than $20$. — Jyrki Lahtonen, Feb 15 '20 at 16:47
See this question. The group is also also denoted by $U(20)$ and is not cyclic, see this duplicate. — Dietrich Burde, Feb 15 '20 at 17:55

score 3 · Answer 1 · 2020-02-15T17:02:34.357

3

$\operatorname{Aut}(\Bbb Z_{20})\cong\Bbb Z_{20}^×\cong(\Bbb Z_5×\Bbb Z_4)^×\cong\Bbb Z_5^××\Bbb Z_4^×\cong\Bbb Z_4×\Bbb Z_2$ and thus is not cyclic.

edited Feb 15 '20 at 17:02

answered Feb 15 '20 at 16:57

What means $G^×$ in your solution? – newuser458 Feb 16 '20 at 20:49
The group of units. – Feb 16 '20 at 20:53

Bernard · Answer 2 · 2020-02-15T21:16:48.033

2

Hint:

An automorphism of $\mathbf Z/20\mathbf Z$ maps the generator $1$ on to another generator, so you have to determine the other generators.

Now for an element $g$ with order $r$ in a commutative group $G$ (in multiplicative notation), $g^k$ has order $\frac r{\gcd(r,k)}$, so for $g^k$ to be another generator, we need to have $k$ and $r$ coprime. In the present case the means that $k$ has to be not divisible by $2$ nor $5$.

There are in all $8$ elements $<20$ coprime to $20$., so the group of automorphism you're after has order $8$. Up to isomorphism, there are exactly $5$ groups of order $8$: the cyclic group $\mathbf Z/8\mathbf Z$, the abelian groups $\mathbf Z/2\mathbf Z\times\mathbf Z/4\mathbf Z$ and $\mathbf Z/2\mathbf Z\times\mathbf Z2\mathbf Z$ and two non-commutative groups: the dihedral group $D_4$ and the quaternionic group $Q_8$.

Can you say which group $\operatorname{Aut}(\mathbf Z/20\mathbf Z)$ is?

edited Feb 15 '20 at 21:16

answered Feb 15 '20 at 16:54

Bernard

175,478

@ChrisCuster: You're right. I just enumerated the elements coprime to 20, and didn't even think of applying the formula to check my count. Thanks for pointing it! – Bernard Feb 15 '20 at 17:25
Using answer from Chris Custer that $Aut (\mathbb Z_{20})$ is not cyclic I can say that I need to exclude cyclical groupsbut without answer Chris I don't know which groups are correct. Can you then develop your statement? – newuser458 Feb 15 '20 at 18:21
1

You can calculate the order of each of the elements and check there's no element of order $8$, it eliminates the cyclic group, and check there does exist elements of order $4; so it eliminates $(\mathbf Z/2\mathbf Z)^3$. If check it is a commutative group, it eliminates the dihedral and quaternionic group. – Bernard Feb 15 '20 at 18:39

score 2 · Answer 3 · answered Feb 15 '20 at 17:02

Multiplication by $-1$ and by $9$ induce automorphisms of $\mathbb Z / 20 \mathbb Z$. Because $$(-1)^2 = 1, \quad 9^2 \equiv 1 \pmod{20} \,,$$ both are involutions. The group $\operatorname{Aut}(\mathbb Z / 20 \mathbb Z)$ contains at least two elements of order $2$, so it cannot be cyclic.

Find elements of a group $\operatorname{Aut}(\mathbb{Z_{20}})$ automorphisms of cyclic group $\mathbb{Z_{20}}$

3 Answers3