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In a problem I am working on, I am asked to find the number of isomorphisms from $\mathbb{Z}_{12}$ to itself using the fact that if $x$ generates a group $G$ and $\phi$ is an isomorphism from $G$ to itself, then $\langle \phi(x) \rangle=G$.

I know that $1,5,7,$ and $ 11$ generate $G$ individually. So I am guessing that either I have to count the number of bijections on a 4 element set or, since $\langle \phi(1) \rangle=G$ and $\phi(k)=\phi(1)^{k\ mod 12}$, count the number of possible ways to map 1 to the set containing $1,5,7,$ and $ 11$.

My initial thought is that since $\phi(1)$ completely determines this whole space we just need to count the number of possible mappings for $1$. If $\phi$, for example, assigns 1 to 5 and 11 to 7, then $\phi$ need not be well defined, right?

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    $\begingroup$ Look at the map that sends $1 \mapsto 5$. Can you tell where it sends every other element? Since $5$ generates the target group, you're good... $\endgroup$
    – John Hughes
    Commented Aug 3, 2016 at 0:14
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    $\begingroup$ Your work looks good, and the answers below should help you complete the question. A more general fact worth knowing is that $\mathrm{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^{\times}$. You can prove this by generalizing the analysis in your post to explicitly write down the isomorphism. $\endgroup$
    – Alex Wertheim
    Commented Aug 3, 2016 at 0:41

4 Answers 4

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Hint - glue together some things you know:

  • You're right that $\phi(1)$ determines all the rest of $\phi$.

  • You seem to know the possible values for $\phi(1)$.

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  • $\begingroup$ Okay, yes I think I was almost there. Since $\phi(1)$ determines the rest of $\phi$, there is only one way to map $1$ to another generator of $\mathbb{Z}_{12}$. Thus, for each possible choice of $\phi(1)$, there is exactly one possible $\phi$. $\endgroup$
    – Mark B
    Commented Aug 3, 2016 at 0:36
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It might be easier to break the problem down into two simpler problems: $$\mathbb{Z}/12\mathbb{Z} \approx \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}.$$ Hence, determining the separate automorphisms of $\mathbb{Z}/3\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$ allows one to glue them together to find the automorphisms of $\mathbb{Z}/12\mathbb{Z}$. For $\mathbb{Z}/3\mathbb{Z}$, one can see that the possibilities are $\varphi(1) = 1$ and $\varphi(1) = 2$. For $\mathbb{Z}/4\mathbb{Z}$, the possibilites are $\varphi(1) = 1$ and $\varphi(1) = 3$. Since there are two automorphisms for each of $\mathbb{Z}/3\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$, these can be glued together to get four automorphisms of $\mathbb{Z}/12\mathbb{Z}$.

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Suppose $\phi$ is an automorphism of the required group. Then we may go through the possible values of $\phi(1)$ and see what happens.

In particular, if you do this and use the assumption that $\phi$ is an isomorphism then you'll see that if $\phi(1)\not\in\{1,5,7,11\}$ then we reach a contradiction: In such a case $\phi$ will not be an isomorphism. For example, if $\phi(1)=2$ then $6\phi(1)=0$, contradicting the facts that the order of the group is $12$ and, if $\phi$ is an isomorphism, then $\langle \phi(1)\rangle=\mathbb{Z}_{12}$.

On the other hand, we can explicitly define automorphisms with $\phi(1)\in\{1,5,7,11\}$ and, furthermore, we can prove that the isomorphism we've defined with, say, $1\mapsto1$ is the only one possible.

Since this exhausts all possibilities, we are done: These four automorphisms are the only ones possible.

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An automorphism is defined by the image of $1$, and it has to map $1$ onto a generator of the group. It is known the generators of $\mathbf Z/n\mathbf Z$ are the (congruences classes of) natural numbers between $0$ and $n$ which are coprime to $n$. There are $\varphi(n)$ (the totient function) of them.

Now $\varphi(12)=\varphi(3)\varphi(4)=2\cdot 2$. Explicitly: $$\{1,5, 7,11\}.$$

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