Groups

Let \(n_q\) be the number of Sylow \(q\)-subgroups. Sylow’s theorems give \[ n_q\mid p, \qquad n_q\equiv1\pmod q. \] Thus \(n_q\in\{1,p\}\). Since \(p<q\), the value \(p\) cannot be congruent to \(1\pmod q\). Hence \(n_q=1\).

Its unique Sylow \(q\)-subgroup has order \(q\), so it is a nontrivial proper normal subgroup of \(G\).

Let \(n_p\) be the number of Sylow \(p\)-subgroups. Sylow’s theorems give \[ n_p\mid q, \qquad n_p\equiv1\pmod p. \] Hence \(n_p\in\{1,q\}\). If \(n_p=q\), then \(q\equiv1\pmod p\), i.e. \(p\mid(q-1)\), contrary to hypothesis. Therefore \(n_p=1\).

Let \(P\) and \(Q\) be the unique Sylow \(p\)- and \(q\)-subgroups. They are normal, \(P\cap Q=1\), and \(|PQ|=pq=|G|\), so \(G=PQ\). For \(x\in P\) and \(y\in Q\), the commutator \([x,y]\) lies in both \(P\) and \(Q\), because both subgroups are normal. Hence \([x,y]=1\), so \(P\) and \(Q\) commute and \(G\cong P\times Q\). Groups of prime order are cyclic, and the direct product of cyclic groups of coprime orders \(p\) and \(q\) is cyclic of order \(pq\).

The cyclic group \(C_{pq}\) always occurs. Since \[ \operatorname{Aut}(C_q)\cong C_{q-1}, \] the divisibility \(p\mid(q-1)\) gives a subgroup of order \(p\) in \(\operatorname{Aut}(C_q)\), hence a nontrivial homomorphism \(C_p\to\operatorname{Aut}(C_q)\) and therefore a nonabelian semidirect product \(C_q\rtimes C_p\). Because \(\operatorname{Aut}(C_q)\) is cyclic, it has a unique subgroup of order \(p\); all nontrivial actions have the same image and yield isomorphic semidirect products. Thus these are exactly the two isomorphism types.

The quaternion group is \(Q_8=\{1,-1,i,-i,j,-j,k,-k\}\), with \(i^2=j^2=k^2=-1\), \(ij=k\), and \(ji=-k\) [@DF04]. Thus it is nonabelian. By Lagrange’s theorem, a proper subgroup has order \(1\), \(2\), or \(4\). The identity subgroup is normal. The only element of order \(2\) is \(-1\), since each of \(\pm i,\pm j,\pm k\) squares to \(-1\). Hence the only subgroup of order \(2\) is \(\{1,-1\}\), which is central and normal. Every subgroup of order \(4\) has index \(2\) and is normal: for an element outside it, the left and right cosets are both its complement. This covers all proper subgroup orders.

If \(G\) is abelian and \(H\leq G\), then \(ghg^{-1}=h\) for every \(g\in G\) and \(h\in H\). Hence \(gHg^{-1}=H\) for every \(g\), so every subgroup is normal.

For a noncentral element, the size of its conjugacy class is \([G:C_G(g)]\), a power of \(p\) greater than \(1\), so it is divisible by \(p\). Splitting \(G\) into conjugacy classes therefore gives \(|G|\equiv|Z(G)|\pmod p\). The center contains the identity, and its order divides \(p^2\). Consequently \(|Z(G)|\) is \(p\) or \(p^2\).

If \(|Z(G)|=p^2\), then \(G=Z(G)\) is abelian. Otherwise \(G/Z(G)\) has prime order and is cyclic. Choose \(g\in G\) whose coset generates this quotient. Every element of \(G\) is \(g^a z\) for some integer \(a\) and \(z\in Z(G)\). Two such elements commute, because their powers of \(g\) commute and their central factors commute with everything. Thus \(G\) is abelian in this case as well.

We prove by induction on \(|G|\) that every finite abelian group has a subgroup of each positive order dividing \(|G|\). For \(d=1\), the identity subgroup works; this includes the base case \(|G|=1\).

Suppose \(d>1\) and choose a prime \(p\mid d\). Since \(p\mid|G|\), Cauchy’s theorem supplies an element of order \(p\), generating a subgroup \(N\) of order \(p\) [@DF04]. It is normal because \(G\) is abelian. The quotient \(G/N\) is abelian of order \(|G|/p<|G|\), and \(d/p\) divides its order. By induction it has a subgroup \(\overline H\) of order \(d/p\). The inverse image of \(\overline H\) in \(G\) is a subgroup consisting of \(d/p\) cosets of \(N\), each of size \(p\). Its order is \(d\), as required.

This is the group of sequences \((a_1,a_2,\ldots)\) with \(a_n\in\mathbb Z/2\mathbb Z\) and only finitely many nonzero coordinates, under coordinatewise addition. It is infinite because the sequences having a single \(1\) in position \(n\) are distinct as \(n\) varies. Every element added to itself is zero, so every element has order \(1\) or \(2\), and in particular finite order.

This group has order \(8=2^{2+1}\cdot1\), with \(2\nmid1\). The subgroup \(H_1=\langle(1,0)\rangle\) has order \(4\) and is cyclic. The subgroup \[ H_2=\langle(2,0),(0,1)\rangle =\{(0,0),(2,0),(0,1),(2,1)\} \] also has order \(4=2^k\), but every nonidentity element has order \(2\). Thus \(H_2\cong C_2\times C_2\) and \(H_1\cong C_4\) are not isomorphic, since an isomorphism preserves element orders.

setup.

\(|G| = 5 > 1\).

Lagrange’s theorem.

\(5\) is prime, so its only positive divisors are \(1\) and \(5\).

<2>4. <2>6. No nonabelian group of order \(5\) exists.

<2>5.

order formula for symmetric groups on \(3\) elements.

definitions of transpositions in \(S_3\).

direct cycle composition: \(1 \mapsto 1 \mapsto 2\), \(2 \mapsto 3 \mapsto 3\), \(3 \mapsto 2 \mapsto 1\).

direct cycle composition: \(1 \mapsto 2 \mapsto 3\), \(2 \mapsto 1 \mapsto 1\), \(3 \mapsto 3 \mapsto 2\).

\((1\ 2\ 3) \neq (1\ 3\ 2)\) (for example, they map \(1\) to different elements).

<2>1 and <2>5.

<1>1 and <1>2.

The sign homomorphism \(S_4\to\{1,-1\}\) is surjective because a transposition has sign \(-1\). Its kernel \(A_4\) therefore has order \(4!/2=12\) [@DF04]. For a \(3\)-cycle, choose the three letters in four ways and then one of the two cyclic orientations. This gives exactly \(4\cdot2=8\) distinct \(3\)-cycles. Each is even, since \((a\,b\,c)=(a\,c)(a\,b)\), so all eight lie in \(A_4\). They all have order \(3\).

If \(H\leq A_4\) had order \(6\), its index would be \(2\). Every index-two subgroup is normal: for an element outside \(H\), both its left coset and its right coset are the complement of \(H\), while an element of \(H\) has both cosets equal to \(H\). Thus the quotient map \(\pi:A_4\to A_4/H\) would be defined, with target of order \(2\).

For a \(3\)-cycle \(c\), the order of \(\pi(c)\) divides both \(3\) and \(2\), so \(\pi(c)=1\). Hence all eight distinct \(3\)-cycles from step <1>1 would belong to \(H\). Together with the identity they would give at least nine elements in a group of order six, a contradiction.

Lagrange’s theorem asserts that a subgroup order divides the group order. Here that divisibility holds for \(6\mid12\) but the requested subgroup does not exist, disproving the converse.

If \(g^a=g^b\), then \(n\mid a-b\), so \(n\mid ka-kb\) and \(g^{ka}=g^{kb}\); thus \(\phi_k\) is well defined, and \(\phi_k(g^ag^b)=g^{k(a+b)}=\phi_k(g^a)\phi_k(g^b)\). An endomorphism \(\phi\) is determined by \(\phi(g)\), which equals \(g^k\) for some \(k\); then \(\phi=\phi_k\). Since \(\phi_k(g)=g^k\), the map \(\phi_k\) depends only on \(k\bmod n\).

Since \(C_n\) is finite, \(\phi_k\) is bijective if and only if it is surjective, that is, if and only if \(g^k\) generates \(C_n\). The order of \(g^k\) is \(n/\gcd(n,k)\), which equals \(n\) exactly when \(\gcd(n,k)=1\).

By steps <1>1 and <1>2, \(\Psi\) is well defined. For units \(k,\ell\), \((\phi_k\circ\phi_\ell)(g)=\phi_k(g^\ell)=g^{k\ell}=\phi_{k\ell}(g)\), so \(\phi_k\circ\phi_\ell=\phi_{k\ell}\) and \(\Psi\) is a homomorphism. If \(\phi_k=\operatorname{id}\), then \(g^{k-1}=e\) and \(k\equiv1\pmod n\), so \(\Psi\) is injective. By steps <1>1 and <1>2, every automorphism equals \(\phi_k\) with \(\gcd(k,n)=1\), so \(\Psi\) is surjective.

Step <1>3 gives \(\Aut(C_n)\cong(\ZZ/n\ZZ)^\times\).

Normality gives \(gKg^{-1}=K\). Hence \(gPg^{-1}\) is a subgroup of \(K\). Conjugation preserves order, so it has order \(|P|\), the largest power of \(p\) dividing \(|K|\). It is therefore a Sylow \(p\)-subgroup of \(K\). Sylow conjugacy, applied inside the finite group \(K\), gives an element \(k\in K\) with the asserted equality [@DF04].

Choose \(k\) as in step <1>1. Then \[ (k^{-1}g)P(k^{-1}g)^{-1} =k^{-1}(gPg^{-1})k=P. \] Thus \(k^{-1}g\in N_G(P)\) and \(g=k(k^{-1}g)\in KN_G(P)\). This proves \(G\subseteq KN_G(P)\). The reverse inclusion holds because both factors are subsets of the group \(G\) and their products lie in \(G\). Hence \(G=KN_G(P)\).

Conjugation is transitive on the Sylow subgroups by step <1>1, and the stabilizer of \(P\) is its normalizer \(N_G(P)=\{g\in G:gPg^{-1}=P\}\). Orbit-stabilizer gives the index formula [@DF04]. Since \(P\leq N_G(P)\), \[ m=[G:P]=[G:N_G(P)]\,[N_G(P):P], \] so \(n_p\mid m\).

For the congruence, restrict the conjugation action to \(P\). Every orbit has power-of-\(p\) size by orbit-stabilizer; hence all non-singleton orbits have size divisible by \(p\). The subgroup \(P\) itself is a fixed point. If another Sylow subgroup \(Q\) is fixed, then \(P\) normalizes \(Q\). Consequently \(PQ\) is a subgroup, of order \(|P||Q|/|P\cap Q|\), a power of \(p\). Its order divides \(|G|\) and hence is at most \(p^a\). But it contains both order-\(p^a\) groups \(P,Q\), forcing \(P=PQ=Q\). Thus there is exactly one fixed point, and counting orbits gives \(n_p\equiv1\pmod p\).

If \(P\) is normal, its conjugacy orbit has one member; conjugacy in step <1>1 then implies that it is the only Sylow subgroup. Conversely, if it is unique, each conjugate has the same order and must equal \(P\), which is normality. For \(a=0\), the unique Sylow \(p\)-subgroup is the trivial group.

The double transpositions \((1\,2)(3\,4)\), \((1\,3)(2\,4)\), \((1\,4)(2\,3)\) are even, commute, and together with \(1\) form the Klein four-group \(V_4\) of order \(4\). A \(3\)-cycle and a \(5\)-cycle are even and generate cyclic subgroups of orders \(3\) and \(5\).

The index of the normalizer of a Sylow subgroup is the number of Sylow subgroups for that prime [@DF04], so it suffices to count Sylow subgroups and exhibit a subgroup of the right order inside each normalizer. The \(15\) double transpositions of \(A_5\) lie three to a Klein four-group, so \(n_2=5\) and \(\abs{N_{A_5}(P_2)}=12\); since \(V_4\lhd A_4\), the normalizer is \(A_4\). The \(3\)-cycles in \(A_5\) number \(20\), so \(n_3=10\) and \(\abs{N_{A_5}(P_3)}=6\); conjugation by \((1\,2)(4\,5)\) sends \((1\,2\,3)\) to \((2\,1\,3)=(1\,2\,3)^{-1}\), so the displayed group of order \(6\) normalizes \(P_3\), and it is nonabelian, hence \(\cong S_3\). The \(5\)-cycles number \(24\), so \(n_5=6\) and \(\abs{N_{A_5}(P_5)}=10\); conjugation by \((2\,5)(3\,4)\) sends \((1\,2\,3\,4\,5)\) to \((1\,5\,4\,3\,2)=(1\,2\,3\,4\,5)^{-1}\), so the displayed dihedral group of order \(10\) is the normalizer.

The normalizers of \(P_3\) and \(P_5\) in step <1>2 have orders \(6\) and \(10\). A group of order \(15\) is cyclic, because its Sylow \(3\)- and \(5\)-subgroups are both unique and hence normal, so it is their direct product \(\ZZ/3\times\ZZ/5\cong\ZZ/15\). The elements of \(A_5\) have cycle types \(1^5\), \(2^21\), \(31^2\), \(5\), hence orders \(1,2,3,5\); none has order \(15\).

Steps <1>1, <1>2, and <1>3 answer parts (a), (b), and (c).

Existence, conjugacy, divisibility, and the congruence are the Sylow theorems [@DF04]. The bound \(n_p\leq m\) follows from the positive divisibility \(n_p\mid m\). Conjugation acts transitively on the Sylow subgroups, and the stabilizer of \(P\) is its normalizer \(N_G(P)\); the orbit-stabilizer formula gives the displayed index.

A unique Sylow subgroup is fixed by all conjugations, so is normal. Conversely, normality makes its conjugacy orbit a singleton, and Sylow conjugacy then gives uniqueness. Every automorphism of \(G\) preserves subgroup orders and therefore permutes the Sylow \(p\)-subgroups; a unique one is consequently characteristic. Finally, a positive integer larger than \(1\) and congruent to \(1\) modulo \(p\) is at least \(p+1\).

Suppose \(H<G\) has finite index \(n\leq4\). Properness implies \(2\leq n\leq4\). The action on the left cosets defines a homomorphism \[ \rho:G\longrightarrow\operatorname{Sym}(G/H)\cong S_n, \qquad \rho(g)(aH)=gaH. \] Its kernel is a normal subgroup of \(G\). It is not all of \(G\): if \(g\notin H\), then \(\rho(g)\) moves the coset \(H\) to the different coset \(gH\). Simplicity therefore makes the kernel trivial.

Thus \(G\) embeds in \(S_n\), so it is finite and \(|G|\leq n!\leq24\). Since \(A_5\) is the smallest nonabelian simple group, \(|G|\geq|A_5|=5!/2=60\), a contradiction.

Since \(90=2\cdot3^2\cdot5\), the structure theorem for finite abelian groups gives a unique decomposition into cyclic groups of prime-power order whose exponents for each prime \(p\) form a partition of the exponent of \(p\) in \(90\) [@DF04]. The \(2\)-part and \(5\)-part are \(\ZZ/2\) and \(\ZZ/5\); the \(3\)-part is \(\ZZ/9\) or \(\ZZ/3\times\ZZ/3\). By the Chinese remainder theorem, \(\ZZ/2\times\ZZ/9\times\ZZ/5\cong\ZZ/90\) and \(\ZZ/2\times\ZZ/3\times\ZZ/3\times\ZZ/5\cong\ZZ/3\times\ZZ/30\).

Over \(\RR\), \(x^4-1=(x-1)(x+1)(x^2+1)\) with pairwise distinct irreducible factors. The module \(M\) is finitely generated and torsion over the principal ideal domain \(\RR[x]\), so by the structure theorem it is a direct sum of cyclic modules \(\RR[x]/(p^k)\) with \(p\) monic irreducible, and the multiset of these summands is determined by \(M\) [@DF04]. Each \(p^k\) divides \(x^4-1\), which is squarefree, so \(k=1\) and \(p\in\{x-1,x+1,x^2+1\}\). Hence \(M\cong M_{a,b,c}\) for unique \(a,b,c\ge0\), and \(\dim_\RR M=a+b+2c\). The solutions of \(a+b+2c=4\) are the five triples with \(c=0\), the three with \(c=1\) and \(a+b=2\), and \((0,0,2)\). Each \(M_{a,b,c}\) is annihilated by \(x^4-1\) and has dimension \(4\).

Step <1>1 answers part (a) and step <1>2 answers part (b).

If \(\gcd(a,b)=1\), then \(\mathbf{Z}_a\times\mathbf{Z}_b\cong\mathbf{Z}_{ab}\). Hence \(\mathbf{Z}_{15}\cong\mathbf{Z}_3\times\mathbf{Z}_5\), \(\mathbf{Z}_{20}\cong\mathbf{Z}_{2^2}\times\mathbf{Z}_5\), and \[ G\cong\mathbf{Z}_{2^2}\times\mathbf{Z}_3\times\mathbf{Z}_{3^2}\times\mathbf{Z}_5\times\mathbf{Z}_5, \] which lists the elementary divisors. Grouping the largest prime powers gives \(\mathbf{Z}_{2^2}\times\mathbf{Z}_{3^2}\times\mathbf{Z}_5\cong\mathbf{Z}_{180}\), and the rest gives \(\mathbf{Z}_3\times\mathbf{Z}_5\cong\mathbf{Z}_{15}\). Thus \(G\cong\mathbf{Z}_{15}\times\mathbf{Z}_{180}\) with \(15\mid180\), and by uniqueness of invariant factors [@DF04] these are the invariant factors of \(G\).

Since \(2700=2^2\cdot3^3\cdot5^2\), an abelian group of order \(2700\) is determined up to isomorphism by one partition of each exponent, the parts giving the orders of its cyclic primary factors [@DF04]. The exponent \(2\) has the \(2\) partitions \((2),(1,1)\), and the exponent \(3\) has the \(3\) partitions \((3),(2,1),(1,1,1)\). The number of groups is therefore \(2\cdot3\cdot2=12\).

Step <1>1 answers part (a) and step <1>2 answers part (b).

Integer row operations replace the relations by another generating set of the same subgroup, and invertible integer column operations change the basis of \(\ZZ^3\); neither changes the isomorphism type of \(M\). Subtract \(4\) times the first column from the second and the first column from the third, then negate the second row. The rows become \((0,4,2)\), \((1,0,0)\), \((0,0,-2)\). The remaining \(2\times2\) block in the last two columns is \(\begin{pmatrix}4&2\\0&-2\end{pmatrix}\). Subtracting twice its second column from its first gives \(\begin{pmatrix}0&2\\4&-2\end{pmatrix}\); adding its first row to its second gives \(\begin{pmatrix}0&2\\4&0\end{pmatrix}\); swapping its columns gives \(\operatorname{diag}(2,4)\). Since \(1\mid2\mid4\), this is the Smith normal form. As a check, \(\abs{\det A}=8=1\cdot2\cdot4\).

By step <1>1, \(M\cong\ZZ/1\oplus\ZZ/2\oplus\ZZ/4\), and \(\ZZ/1=0\).

Step <1>2 gives the isomorphism type.

In an abelian group \(A\) of order \(60=4\cdot3\cdot5\), each Sylow subgroup is normal and hence unique by Sylow conjugacy [@DF04]. Multiplication from the product of these subgroups to \(A\) is a homomorphism because the factors commute. Its kernel is trivial: if \(uvw=1\) with \(u,v,w\) in the order-\(4\), order-\(3\), and order-\(5\) factors, then \(u=(vw)^{-1}\) has order dividing both \(4\) and \(15\), so \(u=1\). Now \(v=w^{-1}\) has order dividing both \(3\) and \(5\), so \(v=w=1\). The domain and codomain both have order \(60\), proving that multiplication is an isomorphism. The order-\(3\) and order-\(5\) factors are cyclic. An abelian group of order \(4\) either contains an element of order \(4\), in which case it is \(C_4\), or has every nonidentity element of order \(2\), in which case it is a two-dimensional vector space over \(\mathbb F_2\) and is \(C_2\times C_2\).

Thus the possibilities are \(C_4\times C_3\times C_5\cong C_{60}\) and \(C_2\times C_2\times C_3\times C_5 \cong C_2\times C_{30}\). The displayed identifications combine cyclic factors of coprime orders; a tuple of generators has order equal to the product of those orders. The two groups are not isomorphic, since only the first contains an element of order \(4\).

Factor \[ f=x^8-1=(x-1)(x+1)(x^2+1)(x^4+1). \] The quadratic has no rational root and is irreducible. The substitution \(x=y+1\) changes \(x^4+1\) to \(y^4+4y^3+6y^2+4y+2\), which is Eisenstein at \(2\) [@DF04]. Since substitution by \(y+1\) is an invertible change of polynomial variable, \(x^4+1\) is also irreducible. Together with the two linear factors, these are distinct monic irreducibles and are therefore pairwise coprime.

The Chinese remainder theorem gives \[ R=\mathbb Q[x]/(f)\cong E_1\times E_2\times E_4\times E_8 \] [@DF04]. In particular, each \(E_j\) is a field. Their dimensions over \(\mathbb Q\) are respectively \(1,1,2,4\): for a monic polynomial of degree \(r\), polynomial division gives the basis \(1,x,\ldots,x^{r-1}\) in its quotient.

Because \(fM=0\), the action on \(M\) factors through \(R\). Let \(e_j\in R\) be the element whose coordinate in \(E_j\) is \(1\) and whose other coordinates are \(0\). Then \(\sum_j e_j=1\), \(e_j^2=e_j\), and \(e_je_k=0\) for \(j\ne k\). Thus every \(m\in M\) equals \(\sum_j e_jm\). If a sum of elements of the \(e_jM\) is zero, multiplying by \(e_k\) shows that its \(k\)th term is zero. Consequently \(M=\bigoplus_j e_jM\).

Each \(e_jM\) is a vector space over \(E_j\). It is finite-dimensional, since any linearly independent family over \(E_j\) is also linearly independent over \(\mathbb Q\) and \(\dim_{\mathbb Q}M=4\). Choose bases of sizes \(a,b,c,d\). This proves the claimed decomposition and dimension equation. Conversely, every direct sum satisfying that equation has dimension \(4\) and is annihilated by \(f\).

A \(\mathbb Q[x]\)-module isomorphism commutes with the action of each \(e_j\) and hence restricts to an \(E_j\)-linear isomorphism on the corresponding summand. It therefore preserves all four multiplicities. Equal multiplicities give isomorphic direct sums, proving uniqueness.

The equation in step <1>2 implies \(d\leq1\). For \(d=1\), the other multiplicities vanish, giving the last row. For \(d=0\), one has \(c\leq2\). If \(c=2\), then \(a=b=0\). If \(c=1\), then \(a+b=2\), giving the three pairs \((2,0),(1,1),(0,2)\). If \(c=0\), then \(a+b=4\), giving the five pairs displayed in the first five rows. These cases are exhaustive and disjoint, and uniqueness in step <1>2 shows that no two rows are isomorphic.

An abelian group is a \(\ZZ\)-module, and \(\ZZ\) is a principal ideal domain; a group of order \(72\) is finitely generated with \(r=0\), so step <1>1 applies with \(R=\ZZ\). Since \(72=2^3\cdot3^2\), the elementary-divisor form writes \(G\cong G_2\oplus G_3\) with \(\abs{G_2}=8\) and \(\abs{G_3}=9\), and the isomorphism types of \(G_2\) and \(G_3\) correspond to the partitions \(3\), \(2+1\), \(1+1+1\) of \(3\) and \(2\), \(1+1\) of \(2\). The \(3\cdot2=6\) combinations are the rows of the table; the invariant factors in each row are obtained by combining the largest \(2\)-power with the largest \(3\)-power, and so on, by the Chinese remainder theorem.

A matrix \(A\in M_3(\QQ)\) makes \(V=\QQ^3\) a finitely generated torsion module over the principal ideal domain \(\QQ[x]\) with \(x\cdot v=Av\), and two matrices are conjugate exactly when these modules are isomorphic. By step <1>1, \(V\cong\QQ[x]/(f_1)\oplus\cdots\oplus\QQ[x]/(f_k)\) with monic \(f_1\mid\cdots\mid f_k\) and \(f_1\cdots f_k=\det(xI-A)=x(x-1)^2\). Since each \(f_i\) divides \(f_k\), a factor \(x\) appearing in \(f_1\) would appear twice in the product; so \(x\) divides only \(f_k\), and either \(k=1\) with \(f_1=x(x-1)^2\), or \(k=2\) with \(f_1=x-1\) and \(f_2=x(x-1)\). The rational canonical forms are the block diagonal matrices of the companion matrices of the invariant factors.

Step <1>1 states the theorem for part (a). Steps <1>2 and <1>3 answer parts (b) and (c) as the cases \(R=\ZZ\) and \(R=\QQ[x]\) of that theorem.

For three pairs, multiplying the first two and then the third gives \[ \bigl(j\varphi_\ell(j')\varphi_{\ell\ell'}(j''), \ell\ell'\ell''\bigr). \] Multiplying the last two and then the first gives \[ \bigl(j\varphi_\ell(j'\varphi_{\ell'}(j'')), \ell\ell'\ell''\bigr). \] These are equal because \(\varphi_\ell\) preserves multiplication and \(\varphi_{\ell\ell'}=\varphi_\ell\circ\varphi_{\ell'}\). Thus the operation is associative.

The identity is \((1_J,1_L)\), since \(\varphi_{1_L}=\operatorname{id}_J\) and each automorphism fixes \(1_J\). The inverse of \((j,\ell)\) is \[ \bigl(\varphi_{\ell^{-1}}(j^{-1}),\ell^{-1}\bigr). \] Multiplication in the first order gives first coordinate \(j\varphi_\ell\varphi_{\ell^{-1}}(j^{-1})=1_J\); in the reverse order it gives \(\varphi_{\ell^{-1}}(j^{-1})\varphi_{\ell^{-1}}(j)=1_J\). Both second coordinates are \(1_L\). This verifies all group axioms.

If every \(\varphi_\ell\) is the identity on \(J\), the multiplication in step <1>1 becomes \((j,\ell)*(j',\ell')=(jj',\ell\ell')\), the direct-product multiplication. The identity function is then an isomorphism.

Conversely, suppose this identity function preserves multiplication. For any \(j\in J\) and \(\ell\in L\), multiplication in the semidirect product gives \[ (1_J,\ell)*(j,1_L)=(\varphi_\ell(j),\ell), \] whereas multiplication of the same pairs in the direct product gives \((j,\ell)\). Preservation by the identity map implies \(\varphi_\ell(j)=j\) for every \(j,\ell\). Thus every \(\varphi_\ell\) is the identity automorphism, proving the converse.

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.

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 \(\langle\varphi\rangle\cong C_n\).

Every subgroup of a cyclic group is cyclic [@DF04], so step <1>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 <1>2 gives \(\operatorname{Aut}(F_n)\cong C_n\), realizing the prescribed cyclic group. For \(n=1\) the group is trivial.

Direct substitution gives \[ \sigma(x)=\frac{x-1}{x},\qquad \sigma^2(x)=1-\frac{x}{x-1}=\frac1{1-x},\qquad \sigma^3(x)=1-(1-x)=x. \] The rational functions used here have nonzero denominators because \(x\) is transcendental. Also \(\sigma(x)\ne x\): equality would imply \(x^2-x+1=0\), a nonzero polynomial relation over \(K\). Since the automorphism fixes \(K\) and \(K(x)\) is generated by \(x\), it follows that \(\sigma^3=1\) and \(\sigma\ne1\). Its order is therefore exactly \(3\). In particular \(x,\sigma(x),\sigma^2(x)\) are distinct.

The sum of the three iterates is fixed by \(\sigma\), which cyclically permutes its summands. Combining the fractions gives \[ x+\frac{x-1}{x}+\frac1{1-x} =\frac{x^2(x-1)+(x-1)^2-x}{x(x-1)} =\frac{x^3-3x+1}{x^2-x}=u. \] Thus \(K(u)\subseteq K(x)^G\). Multiplying the formula for \(u\) by \(x^2-x\) and rearranging gives \[ x^3-ux^2+(u-3)x+1=0. \] This is a monic cubic relation. Hence \([K(x):K(u)]\leq3\).

Put \(E=K(x)^G\). Since \(K(u)\subseteq E\subseteq K(x)\), step <1>2 makes \(K(x)/E\) finite of degree at most \(3\). Let \(m(T)\) be the minimal polynomial of \(x\) over \(E\). Every element of \(G\) fixes its coefficients, so applying \(1,\sigma,\sigma^2\) to \(m(x)=0\) shows that \(x,\sigma(x),\sigma^2(x)\) are roots of \(m\). Step <1>1 shows that they are distinct; consequently \(\deg m\geq3\). As \(E\) contains \(K\), one has \(E(x)=K(x)\), and therefore \([K(x):E]=3\).

The tower law now gives \[ 3\geq[K(x):K(u)]=[K(x):E][E:K(u)]=3[E:K(u)]. \] Thus \([E:K(u)]=1\), so \(E=K(u)\) and \([K(x):K(u)]=3\). The monic cubic from step <1>2 must be the minimal polynomial. Its roots are the three distinct iterates above, so it is separable as well. The argument never divides by \(2\) or \(3\) and applies in every characteristic.