# Proof of $\operatorname{Aut}(C_q)\cong C_{q-1}$ by group action?

Let $$G$$ and $$H$$ be groups, and $$\varphi\colon G\to\operatorname{Aut}(H)$$ a homomorphism. Then, further than all the results valid for a general group action on a set, the following additional one holds: $$\operatorname{Fix}(g):=\{h\in H\mid \varphi_g(h)=h\}\le H \tag 1$$ For finite $$G$$ and $$H$$, the condition $$(1)$$ brings new opportunities, as now $$\left|\operatorname{Fix}(g)\right|$$ divides $$|H|$$; as an example, for $$p, q$$ distinct primes such that $$p\nmid q-1$$, one can prove from here that there are no nontrivial homomorphisms $$\phi\colon C_p\to\operatorname{Aut}(C_q)$$, by no means knowing anything on the structure of $$\operatorname{Aut}(C_q)$$. If, in addition, $$\varphi$$ is injective, then this setting may lead to some other result: for instance, for $$q$$ prime and $$H=C_q$$, I was able to prove via $$(1)$$ that $$|G|$$ divides $$q-1$$, again by no means knowing anything on the structure of $$\operatorname{Aut}(C_q)^\dagger$$.

Question. One step ahead in this escalation, if $$\varphi$$ is an isomorphism, I expect this setting to lead to further group action conditions (involving stabilizers, fixed points subgroups, orbits, etc.), potentially useful in getting the structure (isomorphism class) of $$\operatorname{Aut}(H)$$, for some known $$H$$. In particular, as entry test of this approach, I'm aiming to retrieve in this way the known result $$\operatorname{Aut}(C_q)\cong C_{q-1}$$.

Edit (2021-11-5). The integer $$|G|-k$$ is the number of $$g\in G$$ such that $$\left|\operatorname{Fix}(g)\right|=q$$, namely the number of $$g\in G$$ such that $$\varphi_g(h)=h$$ for every $$h\in C_q$$, namely the number of $$g\in G$$ such that $$\varphi_g=\operatorname{Id}_{C_q}$$, whence: $$|G|-k=\left|\operatorname{ker}\varphi\right| \tag{1bis}$$ If $$\varphi$$ is injective, then $$\left|\operatorname{ker}\varphi\right|=1$$ and $$(1\text{bis})$$ yields: $$k=|G|-1$$. But $$k$$ is the number of $$g\in G$$ such that $$\left|\operatorname{Fix}(g)\right|=1$$, namely the number of $$g\in G$$ such that $$\varphi_g(h)=h\Longrightarrow h=1$$. Therefore, every nontrivial $$g\in G$$ is sent to an automorphism of $$C_q$$ which moves all the nontrivial elements of $$C_q$$. If, in addition, $$G\cong\operatorname{Aut}(C_q)$$, then every $$\psi\in\operatorname{Aut}(C_q)\setminus\{\operatorname{Id}_{C_q}\}$$ moves all the nontrivial elements of $$C_q$$. So, $$C_q$$ has at most $$q-1$$ automorphisms, all but one (the identity) of which move all the nontrivial elements of $$C_q$$. Suppose we have proved that there are precisely $$q-1$$ automorphisms: would the fact that $$q-2$$ of them move all the nontrivial elements of $$C_q$$ imply that some of them (automorphisms) must have order $$q-1$$?

Edit (2021-11-8). By the previous edit, every $$\psi\in\operatorname{Aut}(C_q)\setminus\{\operatorname{Id}_{C_q}\}$$ is of the form: \begin{alignat}{1} &\psi(1)=1 \\ &\psi(a^i)=a^{\sigma(i)} \\ \end{alignat} where $$\sigma\in S_{q-1}$$ has cycle type $$(r_1,\dots,r_N)$$ ($$r_1\le\dots\le r_N$$), for some $$N\ge 1$$, with:

• $$r_i\ge 2$$, for every $$i=1,\dots,N$$
• $$\sum_{i=1}^Nr_i=q-1$$

Note that, if a cycle $$(i_1\dots i_s)$$ which composes $$\sigma$$ is such that $$\sum_{j=1}^si_j\not\equiv 0\pmod q$$, then: \begin{alignat}{1} \psi(a^{i_1}\dots a^{i_s}) &= \psi(a^{i_1+\dots+i_s\pmod q}) \\ &= a^{\sigma(i_1+\dots+i_s\pmod q)} \\ \end{alignat} and: \begin{alignat}{1} \psi(a^{i_1})\dots\psi(a^{i_s}) &= a^{\sigma(i_1)}\dots a^{\sigma(i_s)} \\ &= a^{\sigma(i_1)+\dots+\sigma(i_s)\pmod q} \\ \end{alignat} whence: \begin{alignat}{1} a^{\sigma(i_1+\dots+i_s\pmod q)} &= a^{\sigma(i_1)+\dots+\sigma(i_s)\pmod q} \\ \end{alignat} and finally (being the $$a^i$$'s distinct and by definition of cycle): \begin{alignat}{1} \sigma(i_1+\dots+i_s\pmod q) &= \sigma(i_1)+\dots+\sigma(i_s)\pmod q \\ &= i_1+\dots+i_s\pmod q \\ \end{alignat} which is a contradiction, because $$\sigma$$ doesn't fix any element of $$\{1,\dots,q-1\}$$. Therefore, every cycle $$(i_1\dots i_s)$$ which composes $$\sigma$$ must fulfil the condition: $$\sum_{j=1}^si_j\equiv 0\pmod q\tag{1ter}$$ Maybe , the condition $$(1\text{ter})$$, which is fulfilled in particular by the $$(q-1)$$-cycles, limits the possibilities enough that, for some of the nontrivial automorphisms $$\psi_1,\dots,\psi_{q-2}$$, the corresponding $$\sigma$$ must be a $$(q-1)$$-cycle?

Edit (2021-11-29). By taking as known that$$^{\dagger\dagger}$$ $$\operatorname{Aut}(C_q)\cong (\Bbb Z/q\Bbb Z)^\times$$, we can take advantage of $$G$$ being a finite abelian group, and use the framework in the previous Edit to prove, e.g., that $$\operatorname{Aut}(C_7)$$ (and then $$(\Bbb Z/7\Bbb Z)^\times$$) is cyclic. In fact, the order of every element divides the maximal order among the elements in $$G$$. By contradiction, let's assume that such maximal order is $$3$$. Then, all the nontrivial elements must have order $$3$$. But then, the only nontrivial permutations fulfilling the constraints in the previous Edit are: $$\sigma_1=(124)(356)$$, $$\sigma_2=(142)(356)$$, $$\sigma_3=(124)(365)$$, $$\sigma_4=(142)(365)$$, namely too few to build up the whole $$\operatorname{Aut}(C_7)$$. Likewise, let's assume that such maximal order is $$2$$. Then, all the nontrivial elements must have order $$2$$. But then, the only nontrivial permutation fulfilling the constraints in the previous Edit is $$\sigma_1=(16)(25)(34)$$, definitely too few to build up the whole $$\operatorname{Aut}(C_7)$$. Therefore, $$\operatorname{Aut}(C_7)$$ (or, equivalently, $$(\Bbb Z/7\Bbb Z)^\times$$) must have an element of (maximal) order $$6$$. Maybe this kind of argument can be generalized to every $$q$$?

$$^\dagger$$Every automorphism of $$C_q$$ is a permutation of its $$q-1$$ nontrivial elements; therefore, $$\operatorname{Aut}(C_q)\cong K\le S_{q-1}$$ and hence $$\left|\operatorname{Aut}(C_q)\right|$$ divides $$(q-1)!$$. For an embedding $$\varphi\colon G\hookrightarrow\operatorname{Aut}(C_q)$$, $$|G|$$ divides $$\left|\operatorname{Aut}(C_q)\right|$$ and $$\operatorname{Fix}(g):=\{h\in C_q\mid \varphi_g(h)=h\}$$ is a subgroup of $$C_q$$. But $$C_q$$ has no nontrivial subgroups; therefore: \begin{alignat}{1} \sum_{g\in G}\left|\operatorname{Fix}(g)\right| &= k+(|G|-k)q \\ &= |G|q-k(q-1) \\ \tag 2 \end{alignat} for some $$k$$, $$0\le k\le |G|$$. By Burnside's Lemma, $$|G|$$ divides the LHS of $$(2)$$, and hence $$|G|$$ divides $$k(q-1)$$ either. The case $$k=|G|$$ corresponds to a transitive action, whence $$q\mid |G|$$: contradiction, because $$|G| \mid\left|\operatorname{Aut}(C_q)\right|$$ and $$\left|\operatorname{Aut}(C_q)\right|\mid(q-1)!$$, but $$q\nmid (q-1)!$$. Therefore, $$k<|G|$$ and then necessarily $$|G|\mid q-1$$.

$$^{\dagger\dagger}$$After all, here the focus is to prove the cyclicity of $$\operatorname{Aut}(C_q)$$, and the standard result $$\operatorname{Aut}(C_q)\cong U_q$$ can be then assumed as known - see e.g. Herstein's Topics in Algebra, 2nd Edition, Example 2.8.1, page 69.

• OK, here's my worry with your approach. How are you going to use the fact that it's an isomorphism? You seem to want to use the fact that any other permutation of the points that looks like an automorphism is explicitly not allowed to be one, but I don't think there's any obvious test for such things for outer automorphisms. Your idea about embedding in a symmetric group is close to how these things were originally done by Burnside, and when the automorphisms were terms inner and outer in the first place. Aug 13, 2021 at 10:59