What finite groups \(G\) can occur as collections of automorphisms of some field \(F\)? More precisely, for which \(G\) does there exist a field \(F\) such that \(G\) is a subgroup of the automorphism group of \(F\)? Explain.
What finite groups \(G\) can occur as collections of automorphisms of some finite field \(F\)? Explain.
Solution
In part (a), every finite group occurs. In part (b), exactly the finite cyclic groups occur, including the trivial group.
1. Every finite group embeds in the automorphism group ofa field.
Given a finite group \(G\), take algebraically independent indeterminates \(X_h\) indexed by \(h\in G\), and put \(F=\mathbb Q(X_h:h\in G)\). For \(g\in G\), the permutation of the indeterminates given by \[ \sigma_g(X_h)=X_{gh} \] extends to a polynomial-ring automorphism fixing \(\mathbb Q\), with inverse \(\sigma_{g^{-1}}\). It extends further to the fraction field by \(\sigma_g(A/B)=\sigma_g(A)/\sigma_g(B)\): the denominator remains nonzero, and equal fractions remain equal after applying a ring automorphism.
On every variable one has \(\sigma_g\sigma_k(X_h)=X_{gkh}=\sigma_{gk}(X_h)\). Thus \(g\mapsto\sigma_g\) is a homomorphism to \(\operatorname{Aut}(F)\). If \(g\ne1\), then \(\sigma_g(X_1)=X_g\ne X_1\), so it is not the identity. This proves injectivity and realizes \(G\) as a subgroup.
2. The full automorphism group of any finite field is cyclic.
Let \(F\) be finite. Its characteristic is a prime \(p\): the additive order of \(1\) is finite and exceeds one, and a composite order would give zero divisors by factoring it. The prime subfield is therefore \(\mathbb F_p\). If \([F:\mathbb F_p]=n\), then counting coordinates in a basis gives \(|F|=p^n\), with \(n\geq1\).
The Frobenius map \(\varphi(a)=a^p\) is a field homomorphism: the binomial theorem in characteristic \(p\) gives additivity, and multiplication is preserved. It is injective and hence surjective on the finite field, so is an automorphism. Since \(F^\times\) has order \(p^n-1\), Lagrange’s theorem gives \(a^{p^n}=a\) for every \(a\in F\), including zero [@DF04]. Thus \(\varphi^n=1\). If \(0<d<n\) and \(\varphi^d=1\), all \(p^n\) elements would be roots of the nonzero polynomial \(T^{p^d}-T\) of degree \(p^d<p^n\). This contradicts the root bound. Hence \(\varphi\) has order exactly \(n\).
Every automorphism of \(F\) fixes \(1\), and consequently fixes the prime subfield. There are at most \(n\) embeddings of \(F\) over \(\mathbb F_p\) into an algebraic closure. Indeed, adjoin a finite set of generators successively. At a step of degree \(d\), an extension of a given embedding is determined by the image of the adjoined element, and that image must be a root of its transformed minimal polynomial. There are at most \(d\) choices. Multiplying these bounds along the tower gives at most the total degree \(n\).
The \(n\) powers of \(\varphi\) are already distinct automorphisms, so they exhaust \(\operatorname{Aut}(F)\). It is therefore \(\langle\varphi\rangle\cong C_n\).
3. Exactly the finite cyclic groups are realized in part (b).
Every subgroup of a cyclic group is cyclic [@DF04], so step 2 proves necessity. To prove existence for every order \(n\geq1\), fix a prime \(p\) and work in an algebraic closure of \(\mathbb F_p\). The polynomial \(T^{p^n}-T\) has derivative \(-1\), so it has exactly \(p^n\) distinct roots there. Their set \(F_n\) is a field: it contains \(0,1\), and for roots \(a,b\), \[ (a-b)^{p^n}=a-b,\qquad (ab)^{p^n}=ab, \qquad (a^{-1})^{p^n}=a^{-1}\quad(a\ne0). \] The first identity follows by iterating the characteristic-\(p\) binomial identity, and the other two by multiplicativity. Thus \(F_n\) is a field of cardinality \(p^n\). Step 2 gives \(\operatorname{Aut}(F_n)\cong C_n\), realizing the prescribed cyclic group. For \(n=1\) the group is trivial.