$\mathrm{Aut}(\mathbb Z_{11})$ is isomorphic to $\mathbb Z_{10}$.

Question

The question is this:

Prove that $\mathrm{Aut}(\mathbb Z_{11})$ is isomorphic to $\mathbb Z_{10}$.

I tried to construct a mapping from $f\colon\mathbb Z_n\to \mathbb Z_n$ and $f([k])=[ka]$ where $f$ is in $\mathrm{Aut}(\mathbb Z_n)$ (I have proven that..) But I still have no idea how to keep think along this idea. Maybe it is not helpful.I am not sure..

I also confused about is that, $\mathrm{Aut}(\mathbb Z_{11})$ is a group that has a bunch of mappings, and $\mathbb Z_{10}$ is a group. How can I have a mapping, that maps from a mapping to a element of group? How to understand about that?

Could you give some hints about this problem or help me better understand, if possible? Thanks!

You should try to prove directly that if $p$ is a prime, then $\mbox{Aut}(\mathbb{Z}p)$ is isomorphic to $\mathbb{Z}{p-1}$. Somehow, it is easier, since it stresses out the key fact that $p$ is prime. — Julien, Apr 10 '13 at 00:09
Sure julien, but it is harder to prove it for general $p$. I would only advice to show $\mathrm{Aut}(\mathbb{Z}/n) \cong (\mathbb{Z}/n)^*$ for general $n$. (Unfortunately again the wrong notation $\mathbb{Z}_n$ here ...) — Martin Brandenburg, Apr 10 '13 at 00:23
@TWeJ735, "How can I have a mapping, that maps from a mapping to a element of group?" -- $Aut(\mathbb Z_11)$ is a group under composition. Do you see why? — Paul Gustafson, Apr 10 '13 at 00:56

score 2 · Answer 1 · answered Apr 10 '13 at 00:11

2

Hint: Let $f: \mathbb Z_{11} \to \mathbb Z_{11}$ be an automorphism.

If $f([1])=[a]$ what is $f([2])$? What about $f([k])$?

Moreover what can $[a]$ be?

if $g[1]=[b]$ is another automorphism, can you see any connection between $f \circ g, a$ and $b$?

answered Apr 10 '13 at 00:11

N. S.

134,609

So f([2])=[2a], and f([k])=[ka].. – y1wj3 Apr 10 '13 at 00:17
And fog([K])=[2kab]? – y1wj3 Apr 10 '13 at 00:20
@TWeJ735 Are you sure about the $2$ in $2kab$? So, $f$ can be "identified" with the corresponding $a$, $b$ with $g$ but then composition of functions becomes multiplication in $\mathbb Z_{11}$..Does this tell you what the isomorphism is? Note that you get multiplication in $\mathbb Z_{11}$ but also that $a$ cannot be anything, it only be ..... – N. S. Apr 10 '13 at 00:24
Sorry..It is supposed to be [kab]..right? – y1wj3 Apr 10 '13 at 00:32
So [a] have to be one of generators of Z11? – y1wj3 Apr 10 '13 at 00:32
@TWeJ735 When is the function $f[k]=[ak]$ an bijection? – N. S. Apr 10 '13 at 01:34
when f is in Aut(Zn)? I am not sure if that is the direction you want me go.. – y1wj3 Apr 10 '13 at 02:06
@TWeJ735 and for which values of $a$ does that happen? – N. S. Apr 10 '13 at 04:23

score 1 · Answer 2 · answered Nov 03 '15 at 13:46

1

We can use following,

$Aut(Z_{n}) \cong U(n)$ for every positive integer $n$,
$U(p^n)\cong Z_{p^n-p^{n-1}}$ for any odd prime $p$ and positive integer $n$.

Therefore, $Aut(Z_{11}) \cong U(11)$ and $U(11) \cong Z_{10}$.

answered Nov 03 '15 at 13:46

Shahnawaz Ahmad

11

score 1 · Answer 3 · answered Apr 10 '13 at 00:18

1

Hint: Find the order of $\operatorname{Aut}(Z_{11})$. Show that an element in it has the same order. It follows that $\operatorname{Aut}(Z_{11}) \cong Z_n$ where $n = \left| \operatorname{Aut}(Z_{11}) \right|$.

answered Apr 10 '13 at 00:18

Ayman Hourieh

40,828

n=10 since there are 10 generators in $\operatorname{Aut}(Z_{11})$ – y1wj3 Apr 10 '13 at 00:39
Then how to use this fact to show $\operatorname{Aut}(Z_{11}) \cong Z_n$? – y1wj3 Apr 10 '13 at 00:44
@TWeJ735 If you find an element $\sigma$ in $\operatorname{Aut}(Z_{11})$ with order $10$, it follows that this element is a generator of $\operatorname{Aut}(Z_{11})$ because the powers $\sigma^j$ are distinct for $0 \le j < n$. From there, it's clear that $\operatorname{Aut}(Z_{11})$ is cyclic. – Ayman Hourieh Apr 10 '13 at 01:10
@TWeJ735 There are not $10$ generators in $\operatorname{Aut}(\mathbb{Z}_{11})$. For example, the identity element does not generate the group. (Indeed, a group of order $10$ contains $10$ elements, including the identity, an element of order $5$ (why?) and an element of order $2$ (why?). These elements cannot generate the whole group on their own.) – user1729 Apr 10 '13 at 12:14

score 1 · Answer 4 · answered Apr 10 '13 at 00:49

1

Hint: An abelian group of order 10 is cyclic.

answered Apr 10 '13 at 00:49

Fundamental Theorem of Finite Abelian Group? – y1wj3 Apr 10 '13 at 01:05
Since 2 and 5 are relatively prime. – y1wj3 Apr 10 '13 at 01:08
Then Z2 x Z5 is isomorphic to Z10? – y1wj3 Apr 10 '13 at 01:12
I have no idea how to relate this to $\operatorname{Aut}(\mathbb Z_11)$ – y1wj3 Apr 10 '13 at 01:12
First you have to show $\operatorname{Aut}(\mathbb{Z}_{11})$ is abelian group of order 10, then use the hint. – Apr 10 '13 at 01:14
Although this is true, I do not see how it is helpful... – user1729 Apr 10 '13 at 10:24

$\mathrm{Aut}(\mathbb Z_{11})$ is isomorphic to $\mathbb Z_{10}$.

4 Answers4