0
$\begingroup$

Let $[a]$ be generator of $\mathbb Z_n$. Do all isomorphisms from $\mathbb Z_n$ to itself consists of the form $\theta_a([k]) = [ka]$?

  • 0
    Do you know that all the elements in $\;\Bbb Z_n\;$ coprime to $\;n\;$ are a *multiplicative* group? – 2017-01-02
  • 0
    @DonAntonio I do know, I believe it is used in Euler theorem proof. I also come to guess that $\text{Aut}( \mathbb Z_n) \simeq \mathbb Z_{\varphi(n)}$ – 2017-01-02
  • 0
    "isomorphism of ${\mathbb Z}_n$" is unclear, because it could mean group isomorphism or ring isomorphism. – 2017-01-02
  • 0
    @DerekHolt what would be the proper way to put it? My question is about group isomorphism. – 2017-01-02
  • 0
    You could put group isomorphisms, or isomorphisms of $({\mathbb Z}_n,+)$. Personally I prefer to use $C_n$ for the cyclic group of order $n$, because ${\mathbb Z}_n$ has too many different meanings. (When $n=p$ is prime, it is also used to denote the ring of $p$-adic integers.) – 2017-01-02

3 Answers 3

3

Since an automorphism must map a generator to a generator, and $\;[m]\in\Bbb Z_n\;$ is a generator iff $\;g.c.d(m,n)=1\;$ , we have if $\;[a]\;$ is a generator, then an automorphism must map $\;[a]\;$ to $\;[ka]\;$ , for some $\;k\in\left(\Bbb Z_n\right)^*\;$ ... This is based in your answer to my comment.

  • 0
    Can we put it this way, let $[k]$ be generator, then $[ka]$ is generator as well? – 2017-01-02
  • 0
    @MulaKoSaag More like "...then $\;[ka]\;$ **must** be a generator as well.", and thus you know that in fact $\;a,k\in(\Bbb Z_n)^*\;$ – 2017-01-02
  • 0
    does $(\mathbb Z_n )^*$ represent generator set in your answer? – 2017-01-02
  • 0
    @MulaKoSaag If you want to consider it that way: it is in fact the set of all units in the ring $\;\Bbb Z_n\;$ , which coincides with the set of generators of the *cyclic additive* group $\;\Bbb Z_n\;$ – 2017-01-02
  • 1
    haha .. I was only asking about the notation. Anyway I get it and thanks – 2017-01-02
3

The automorphisms of $\mathbf Z/n\mathbf Z$ map a generator of $\mathbf Z/n\mathbf Z$, i.e. an element of the multiplicative group $(\mathbf Z/n\mathbf Z)^\times$ to another generator.

  • 0
    is the sentence complete? – 2017-01-02
  • 0
    @Mula Ko Saag: No. I was disrupted on answering, and then I forgot… Thanks for pointing it! – 2017-01-02
3

If $\mathbb{Z}_n$ is the cyclic group of order $n$ then the automorphisms are precisely $\mathbb{Z}_n^{\times}$ which has order $\phi(n)$ where $\phi$ is Euler's totient function.

The automorphisms need to map generators to generators and yes they appear as $a^p$ for some $p$.