2

For $C_n,$ let $g$ is a generator, but except for non-trivial groups, $e$ is not.

So, ignore $e$, hence: $\{\forall i,j\in \{1,\cdots, n-1\}\,| \,\langle g^i \rangle\mapsto \langle g^j \rangle\}.$

All generators are mapped by any other generator, as all are of the same order.

But, how to get number of automorphisms?

jiten
  • 4,524

2 Answers2

2

$U(n)$ denote the multiplicative group of units of the ring $(\mathbb{Z}_n,+,\cdot )$.

Automorphism of a cyclic group can be completely determined by the image of the generators.

Let $f:C_n\to C_n$ be an automorphism and $C_n=\langle a\rangle$ then $|f(a)|=|a|=n$. Hence $C_n=\langle f(a)\rangle$. Since the number of generators of $C_n$ is exactly $\varphi (n)$, you have exactly $\varphi (n)$ choices for $f(a)$.

Therefore, $\operatorname{Aut}(C_n)\cong U(n)$.

commie trivial
  • 984
  • 2
  • 4
  • 14
Sourav Ghosh
  • 12,997
  • 1
    So, ${1,3,5,7,9,11,13,17,19,21,23,29,31,37, 41,43,47, 51,53,57,59,61,63,67,69,71,73,77,79,81, 83,..$ there are $64$ such values. – jiten Jul 20 '22 at 12:34
  • 1
    $\varphi(160) =\varphi(2^55)=\varphi(2^5)\varphi(5)=(2^5-2^4)4=64$ – Sourav Ghosh Jul 20 '22 at 13:31
  • Speaking in non-mathematical terms, your answer states that : number of automorphisms is guided by no. of generators, as the map of a generator must be among generators only. This means that number of automorphisms is guided by the totient function. But, by notation how it is obvious: say for automorphism of a square; have the map $\varphi:C_4\to C_4$ that can be described by $\varphi(e)=r^x, \varphi(r)=r^y, \varphi(r^2)=r^z, \varphi(r^3)=r^w$ for $0\le x,y,z,w \le 3$. Where it is shown by notation that $e$ cannot map to generators? – jiten Jul 20 '22 at 21:25
2

Let $f\colon\mathbb{Z}_n\rightarrow\mathbb{Z}_n$ be a homomorphism, then it is already uniquely defined by $f(1)$ as using the property of a homomorphism yields: $$f(m)=f(\underbrace{1+\ldots+1}_{m\;\text{times}}) =\underbrace{f(1)+\ldots+f(1)}_{m\;\text{times}} =m\cdot f(1).$$ For the group homomorphism $f_k\colon\mathbb{Z}_n\rightarrow\mathbb{Z}_n,f_k(1)=k\;\text{or}\;1\mapsto k$ to be invertible (and therefore be an automorphism), $k$ has to be invertible in $\mathbb{Z}_n$ (the inverse automorphism is then $f_{k^{-1}}$) and $(\mathbb{Z}_n,+,0)\rightarrow(\mathbb{Z}_n^\times,\cdot,1),k\mapsto f_k$ is an isomorphism (the inverse is $f\mapsto f(1)$), which yields: $$\operatorname{Aut}(\mathbb{Z}_n)\cong\mathbb{Z}_n^\times.$$ In particular, using $160=32\cdot 5=2^5\cdot 5$ and using the Chinese remainder theorem (See here) we get: $$\operatorname{Aut}(\mathbb{Z}_{160})\cong\mathbb{Z}_{160}^\times \cong\mathbb{Z}_{32}^\times\times\mathbb{Z}_5^\times \cong\mathbb{Z}_2\times\mathbb{Z}_4\times\mathbb{Z}_8,$$ where $\mathbb{Z}_p^\times\cong\mathbb{Z}_{p-1}$ for an odd prime $p$ and $\mathbb{Z}_{2^k}^\times\cong\mathbb{Z}_2\times\mathbb{Z}_{2^{k-2}}$ were used.

Samuel Adrian Antz
  • 2,134
  • 1
  • 5
  • 19
  • 1
    This answer is not (yet) complete; you are looking for the multiplicatively invertible elements of each of $\mathbb Z_{32}$ and $\mathbb Z_5$, not the full groups. – Mees de Vries Jul 20 '22 at 12:49
  • You're right, I forgot the multiplicative groups and then it can be simplified further. – Samuel Adrian Antz Jul 20 '22 at 12:55
  • 1
    Does your answer follow a different approach than the other one by @LostinSpace? – jiten Jul 20 '22 at 13:04
  • 1
    Sorry, my fault, it's "yes" of course instead of "no". The argument of @LostinSpace uses the order of an element and the lemma that automorphisms preserve it. I answered a bit later with an approach of directly using that automorphisms have to be invertible without using the order. – Samuel Adrian Antz Jul 20 '22 at 13:22
  • How preservation of order implies generators map to generators alone? Say, taking example of a square: the map $\varphi:C_4\to C_4$ can be described by $\varphi(e)=r^x, \varphi(r)=r^y, \varphi(r^2)=r^z, \varphi(r^3)=r^w$ for $0\le x,y,z,w \le 3$. How, it is evident that $e$ must not map to any generator? – jiten Jul 20 '22 at 21:22
  • Also, does your approach builds up by induction, by taking case of all identity maps first? – jiten Jul 20 '22 at 21:29
  • 1
    No, I don't use induction. If you have $\mathbb{Z}n^\times$, you can't conclude anything about $\mathbb{Z}{n+1}^\times$ in general. – Samuel Adrian Antz Jul 22 '22 at 16:37