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Prove that if $p$ is prime, then $\operatorname{Aut}_{\text{Grp}}(C_p)$ is isomorphic to $C_{p-1}$.

Before, this proof you are asked to compute the group of automorphism of $(\mathbb{Z}, +)$. The answer I got was the identity map and $\beta : n \rightarrow -n$.

Can someone explain to me what $C_{p-1}$ is? $\textbf{ANSWER:}$ Cp is the cyclic group of order p.

What is meant by $\operatorname{Aut}_{\text{Grp}}(C_p)$? $\textbf{ANSWER:}$ $\operatorname{Aut}_{\text{Grp}}(C_p)$ is a group, for all $C_{p}$ of all groups.

And perhaps a sketch of the proof?

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  • $\begingroup$ I was reading a website online and updated the question (or more my doubts) $\endgroup$ – Username Unknown Apr 20 '13 at 7:04
  • $\begingroup$ Think about generators of $C_p$. Note that the number of generator of $C_p$ is $\phi(p)$. $\endgroup$ – Babak Miraftab Apr 20 '13 at 7:06
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I know it's been a long time. In case someone is still having troubles with excerice II.4.15 from "Algebra: Chapter 0" by P. Aluffi.

Since $p$ is prime every non-zero element of $C_p$ has order $p$(since order of elements divide order of group). Any automorphism $C_p \rightarrow C_p$ can be determined by where $g^1$ is mapped. It is easy to see that there can be $p-1$ such morphisms with identity morphism being an idendity element. This defines a group of order $p-1$.

One may treat such a group as a multiplicative group modulo $p$ (if it's not clear to you, let $\phi: G\rightarrow G \in Aut_{Grp}(G)$, with generator $g^1$ being mapped to $g^m$, then $\phi^2 = \phi \circ\phi (g) = g^{m^2}$, so it's just multiplication $mm$).

As it was pointed out in comments, in order to conclude $Aut_{Grp}(C_p)$ is cyclic of order $p-1$ we need to show that $(\mathbb{Z}/p\mathbb{Z})^*$ has an element of order $p-1$(making it cyclic).

Let $G = (\mathbb{Z}/p\mathbb{Z})^*, g \in G$ be an element of maximal order. Since the group is abelian the order of every element divides $|g|$. Therefore, for every $h \in G, h^{|g|} = 1$. As we can see the equation $x^{|g|} = 1$ has $p-1$ solutions modulo $p$ (because $p-1$ is group's order). From the theorem that $x^r = 1$ has at most $r$ solutions modulo $p$ we may conclude that $|g|$ is at least $p-1$, it is also at most $p-1$ as the order of group's element is at most the order of group.

Now, suppose the order of $[m]_p$ in $(\mathbb{Z}/p\mathbb{Z})^*$ equals to $p-1$. Then with $\phi$ defined as above $\phi^{p-1} = x^{m^{p-1}} = id$.
Hence, you may conclude that having an element of order $p-1$ makes $Aut_{Grp}(C_p)$ cyclic of order $p-1$ and thus isomorphic to $C_{p-1}$

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    $\begingroup$ While this is correct, the final part about the group being cyclic really needs a further argument. $\endgroup$ – Tobias Kildetoft Aug 3 '17 at 10:34
  • $\begingroup$ It follows from the point of view that $Aut_{Grp}(C_{p})$ is a multiplicative group modulo $p$. For example, my argument shows that if in $ (\mathbb{Z}/p\mathbb{Z})^* $ the order of $[m]_{p}$ is $k$ then $\phi ^k = id_{C_p}$. Proving that $(\mathbb{Z}/p\mathbb{Z})^*$ has an element of order $p-1$ (and hence cyclic of order $p-1$) is another topic. $\endgroup$ – Atin Aug 3 '17 at 11:02
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    $\begingroup$ No, it really is not another topic, since the group being cyclic was part of the original question. $\endgroup$ – Tobias Kildetoft Aug 3 '17 at 11:05

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