5

Find $\text{Aut}(\mathbb{Z})$.

I found an answer on Yahoo, which I don't completely understand.

Recall that an isomorphism of a cyclic group must carry generator to generator. The only generators of $\mathbb{Z}$ are $1$ and $-1$.

<p>Hence the only automorphisms possible are: $f(n)=n$ and $f(n)=-n$. </p>

<p>Therefore $|\text{Aut}(\mathbb{Z})|=2$ and must be isomorphic to $\mathbb{Z}_2$.</p>

Could someone try to explain this to me, maybe formulated a bit otherwise? Especially the first line 'Recall that .. to generator'.

Thanks in advance.

Dr. Heinz Doofenshmirtz
  • 2,360

1 Answers1

12

If $f$ is an automorphism of $\mathbb Z$, then $f(n) = nf(1)$, so $f$ is completely determined by where it sends $1$ (a generator of $\mathbb Z$). Now, if $f(1) = m$, then $f(n) = nf(1) = nm$, so the image of $f$ is the subgroup $m\mathbb Z$. Thus, if we want $f$ to be surjective, then necessarily $f(1) \in \{-1,1\}$.

William Stagner
  • 4,786