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 $-1$.
Hence the only automorphisms possible are: $f(n)=n$ and $f(n)=-n$.
Therefore $|\text{Aut}(\mathbb{Z})|=2$ and must be isomorphic to $\mathbb{Z}_2$.
Could someone try to explain this to me, maybe formulated a bit otherwise? Especially the first line 'Recall that .. to generator'.
Thanks in advance.
$\endgroup$1 Answer
$\begingroup$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 $f(1) \in \{-1,1\}$.
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