Prove $Aut(G) \cong \mathbb{Z}_n^{\times}$ where $G=(\mathbb{Z}_n , +)$.

Question

$Aut(G)$ denotes the set of all automorphisms with respect to composition. $\mathbb{Z}_n^{\times}$ denotes the set of all invertible elements of $\mathbb{Z}_n$ with respect to multiplication.

Please do not offer a full proof as an answer (where is the fun in that?). I am interested only in finding a specific map that works.

My thoughts: Define $\psi : Aut(G) \rightarrow \mathbb{Z}_n^{\times}$ as $\psi ([m])=\phi_m([x])=[mx]$ (a function which is defined in a previous exercise). I am not confident that this works. Should I continue looking for isomorphisms in this direction? Is the opposite direction more fruitful?

Your $\psi$ is correct, except that it goes $\mathbb Z_n^\times \to \mathrm{Aut}(G)$, not the other way around. — Mees de Vries, Oct 24 '16 at 23:22

score 1 · Accepted Answer · answered Oct 24 '16 at 21:18

1

Hints. Since $G$ is cyclic a morphism from $G$ to $G$ is entierely defined by the image of a generator of $G$. If one asks the morphism to be an isomorphism (only the surjectivity is relevant) on what can be sent a generator of $G$?

answered Oct 24 '16 at 21:18

C. Falcon

19,553

Prove $Aut(G) \cong \mathbb{Z}_n^{\times}$ where $G=(\mathbb{Z}_n , +)$.

1 Answers1