I'm just starting an introduction to abstract algebra and I'm currently learning about groups. My textbook says:

"An endomorphism that is also an isomorphism is called an automorphism. . . . The subset of all automorphisms has the structure of a group under composition. We denote that by Aut$(G)$."

Intuitively (and with proof), I understood why Aut$(G)$ has the structure of a group under composition, and to strengthen my concept of homomorphisms in general, I tried to find the complete subset of all automorphisms of simple groups that I've already worked with many times. The first one I tried to find is Aut$(\mathbb{Z})$ under addition.

The only two homomorphisms I could come up with are

\begin{equation} \begin{split} f &: \mathbb{Z}\mapsto\mathbb{Z} \\ &: x \mapsto x \end{split} \end{equation} and \begin{equation} \begin{split} f &: \mathbb{Z}\mapsto\mathbb{Z} \\ &: x \mapsto -x \end{split} \end{equation} I looked around online and I found a StackExchange article where someone mentioned briefly that Aut$(\mathbb{Z})$ under addition has only two elements; the ones I just mentioned. How can I prove this?

  • 1
    $\begingroup$ Hint: Automorphism maps generator to generator $\endgroup$ – Chinnapparaj R May 21 '18 at 8:11

An automorphism $\phi$ of a cyclic group sends a generator to a generator and $\Bbb{Z}$ has two geneartors $1$ and $-1$. So $\phi(1)=1$ or $\phi(1)=-1$. Since $\phi(n.1)=n \phi(1)$. So the first map becomes the identity and latter one maps $x$ to $-x$


Here's a Herstein's nice proof: [Example $2.8.2$]

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