Showing that $\operatorname{aut}(\Bbb Z_6) \simeq \Bbb Z_2$

Question

Okay, I am trying to show that there are only two automorphisms of $\Bbb Z_6$, the set congruences classes modulo 6.

I know that a homomorphism is uniquely determined by how it maps the generator 1. So, suppose that $[1] \rightarrow [k]$ extends to an automorphism of $\Bbb Z_6$, where $[a]$ denotes our congruence class. Since it is surjective, there exists something in the domain that gets mapped to $[1]$; i.e., $[kn] = [1]$ for some integer $n \in \{0,...,5\}$. This implies $kn = 6q + 1$ for some integer $q$, implying that $kn$ is odd, which happens only if both $k$ and $n$ are odd. This means that $k$ could be $1$,$3$, or $5$. Notice that $k=1$ gives us the identity map, $k=3$ contradicts the fact that our map is injective, but the $k=5$ is giving me a little trouble. Specifically, I am having trouble showing $[1] \rightarrow [5]$ extends to an automorphism.

Injectivity: Suppose that $f([n]) = f([m])$, where $f$ is the extension of $[1] \rightarrow [5]$. Then $[5n] = [5m]$ or $5(n-m) = 6q$ for some integer $q$. For some reason I am blanking as to why this implies $n-m$ is divisible by $6$, which is what I need to show injectivity. I am also having trouble with surjectivity.

EDIT: By the way, this comes from Hungerford chapter 1 section 2, for those interested in knowing exactly what theorems I have at the time of this problem.

Answer 1

Your approach here is a bit of a detour. Instead of asking what gets mapped to $[1]$, just continue thinking about where $[1]$ itself maps. Since an isomorphism (which an automorphism is, by definition) preserves all group theoretic properties, in particular we have that the order of $f([1])$ must be the same as the order of $[1]$, which is $6$. The only other element of $\mathbb Z_6$ with order $6$ is $[5]$.

So the only possible automorphisms maps $[1]$ to $[1]$ (producing the identity map) or to $[5]$, and in the latter case you must be looking at $$ f([n]) = [5n] $$ If you can prove that the latter is indeed an automorphism, you're done.

To show that $ 5m\equiv 5n \pmod 6$ implies $m\equiv n\pmod 6$, try multiplying by $5$ on both sides and exploit that $25\equiv 1\pmod 6$.

Then appeal to the general fact that an injective function between finite sets of the same size is automatically surjective.

Answer 2

It's a lot easier if you exploit the fact that $[5]=[-1]$.

In any case, once you have $5(n-m) = 6q$, can you conclude that $n-m$ is divisible by $2$ at least? How about by $3$? Does that get you what you need?

Answer 3

Hint:

The generators of the (additive) group $\mathbf Z/n\mathbf Z$ are the congruence classes of the elements which are coprime to $n$, and a generator maps a generator onto a generator.