Permutation group of a type to Third Sylow theorem

AbbreviationPermutation group of a type#

The permutation group of a set \(X\), also called the symmetric group on \(X\), is the set of all bijections from \(X\) to itself, with composition of maps as the group operation. It is the group of all symmetries of a set with no further structure. By Cayley's theorem, every group is isomorphic to a subgroup of some permutation group, so permutation groups are as general as groups. For a finite set with \(n\) elements the group is written \(S_n\) and is a basic object of combinatorics and algebra.

Definition

A permutation of a set \(X\) is a bijection from \(X\) to itself, that is, a map \(\sigma \colon X \to X\) that has an inverse map \(\sigma^{-1}\) with \(\sigma \circ \sigma^{-1} = \sigma^{-1} \circ \sigma = \mathrm{id}_X\). The permutations of \(X\) form a group, written \(\mathrm{Sym}(X)\), in which the product \(\sigma\tau\) is the composite that applies \(\tau\) first, so that \((\sigma\tau)(x) = \sigma(\tau(x))\). The identity map is the identity element and the inverse map is the group inverse. Equivalently, \(\mathrm{Sym}(X)\) is the group of invertible elements of the monoid of all maps from \(X\) to itself, where a monoid is a set with an associative operation and an identity element, here composition of maps and the identity map.

Up to isomorphism, the group depends only on the size of \(X\): a bijection \(e \colon X \to Y\) induces an isomorphism from \(\mathrm{Sym}(X)\) to \(\mathrm{Sym}(Y)\), sending \(\sigma\) to \(e \circ \sigma \circ e^{-1}\). For this reason one writes \(S_n\) for the permutation group of \(\{1, \dots, n\}\), or of any set with \(n\) elements.

Examples
  • \(S_1\) and \(S_0\), the permutation groups of a one-element set and of the empty set, are trivial. \(S_2\) has two elements, the identity and the map exchanging the two points.
  • \(S_3\) has six elements. In cycle notation, where \((a\ b\ c)\) sends \(a\) to \(b\), \(b\) to \(c\) and \(c\) to \(a\), one computes \((1\ 2)(2\ 3) = (1\ 2\ 3)\) but \((2\ 3)(1\ 2) = (1\ 3\ 2)\). So \(S_3\) is not commutative, and neither is \(S_n\) for any \(n \ge 3\), since it contains a copy of \(S_3\).
  • On the integers, the translation \(n \mapsto n + 1\) is a permutation of infinite order. More generally, the permutation group of an infinite set is always infinite and contains elements of infinite order: an infinite set contains a countably infinite subset, which can be identified with the integers, and the permutation that translates this copy and fixes every other point has infinite order.
Properties
Group actions and Cayley's theorem

The permutation group of \(X\) acts on \(X\) by evaluation, \(\sigma \cdot x = \sigma(x)\). Conversely, any action of a group \(G\) on \(X\) gives a group morphism from \(G\) to \(\mathrm{Sym}(X)\), sending \(g\) to the permutation \(x \mapsto g \cdot x\). When the action is faithful, that is, when only the identity acts as the identity map, this morphism is injective and \(G\) is isomorphic to its image in \(\mathrm{Sym}(X)\). Applied to the action of \(G\) on itself by left multiplication, this is Cayley's theorem.

In Mathlib

Mathlib defines the permutations of a type as its self-equivalences: a permutation is bundled together with its inverse map and the proofs that the two maps undo each other, rather than being a function with a proof of bijectivity. The group structure is defined for every type, finite or not, and multiplication is composition with the right-hand factor applied first. There is no separate definition of \(S_n\); it is the permutation group of the type of natural numbers below \(n\), and statements about finite permutation groups are made for an arbitrary finite type. Results such as the cycle decomposition and the signature assume the type is finite and that equality of its elements is decidable.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibEquiv.Perm
Uses 1 · Used by 10
Statement usesGroup
Used byDerangements Formula · Roth's theorem · Group action · Alternating group · Alternating groups on 5 letters or more are simple · Cayley's theorem · Jordan–Hölder theorem · Decomposition into cycles with disjoint support · Decomposition into transpositions · Signature

DefinitionCayley's theorem#

Cayley's theorem says that every group is isomorphic to a group of permutations: a group \(G\) can be realized inside the permutation group of the set \(G\) itself, by letting each element act on the others through multiplication. The theorem shows that the abstract axioms of a group describe nothing more general than groups of symmetries of a set. It also reduces the study of finite groups, in principle, to subgroups of the symmetric groups \(S_n\).

Statement

For a set \(X\), write \(\operatorname{Sym}(X)\) for the group of all bijections \(X \to X\) under composition. Every group \(G\) is isomorphic to a subgroup of \(\operatorname{Sym}(G)\). The subgroup is the image of the map \(g \mapsto \lambda_g\), where \(\lambda_g(x) = gx\) is left multiplication by \(g\). This embedding is called the left regular representation of \(G\).

If \(G\) is finite of order \(n\), then numbering its elements makes \(\operatorname{Sym}(G)\) a copy of the symmetric group \(S_n = \operatorname{Sym}(\{1, \dots, n\})\), which has \(n!\) elements. So every group of order \(n\) is isomorphic to a subgroup of \(S_n\). In particular there are only finitely many groups of order \(n\) up to isomorphism, since \(S_n\) has only finitely many subgroups.

Proof

Each \(\lambda_g\) is a bijection of \(G\), with inverse \(\lambda_{g^{-1}}\). The associative law gives \(\lambda_{gh}(x) = ghx = \lambda_g(\lambda_h(x))\), so \(g \mapsto \lambda_g\) is a group homomorphism from \(G\) to \(\operatorname{Sym}(G)\). It is injective, because \(\lambda_g(e) = g\) recovers \(g\) from \(\lambda_g\), where \(e\) is the identity. An injective homomorphism maps its domain isomorphically onto its image, and the image of a homomorphism is a subgroup.

The same argument works for any group action of \(G\) on a set \(X\). Each element \(g\) moves the points of \(X\) by a permutation, and these permutations multiply as the elements of \(G\) do. The action is called faithful when only the identity fixes every point, that is, when distinct elements act differently; for a faithful action the map from \(G\) to \(\operatorname{Sym}(X)\) is injective, so \(G\) is isomorphic to its image. Left multiplication is a faithful action of \(G\) on itself.

Examples
  • Cyclic groups. In the cyclic group of order \(n\) generated by \(a\), multiplication by \(a\) sends \(a^k\) to \(a^{k+1}\), an \(n\)-cycle on the \(n\) elements. The group is realized as the subgroup of \(S_n\) generated by one \(n\)-cycle.
  • The Klein four-group. Its elements are \(e, a, b, c\), where each of \(a, b, c\) has order \(2\) and the product of any two distinct ones is the third. Multiplication by \(a\) swaps \(e\) with \(a\) and \(b\) with \(c\). Numbering the elements \(1, 2, 3, 4\) gives the subgroup \(\{\mathrm{id}, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}\) of \(S_4\).
  • Degree of the embedding. The theorem places \(S_3\) inside \(S_6\), though \(S_3\) already acts faithfully on three points. The degree \(n = |G|\) cannot always be lowered. The quaternion group \(\{\pm 1, \pm i, \pm j, \pm k\}\), with \(i^2 = j^2 = k^2 = ijk = -1\), has order \(8\), and every nontrivial subgroup of it contains \(-1\). In a faithful action only the identity fixes every point, so the stabilizers cannot all contain \(-1\); some point has trivial stabilizer, and its orbit has \(8\) points.
Consequences

Letting \(G\) act on the left cosets of a subgroup, rather than on its elements, gives a refinement. Let \(H\) be a subgroup of finite index \(n\). The action of \(G\) on the \(n\) cosets \(gH\) gives a homomorphism \(G \to \operatorname{Sym}(G/H) \cong S_n\), usually not injective, whose kernel is the normal core of \(H\): the elements of \(G\) all of whose conjugates lie in \(H\). The kernel is a normal subgroup contained in \(H\), and the quotient by it embeds in \(S_n\), so the index of the normal core divides \(n!\).

  • For \(n = 2\) the index of the normal core is a multiple of \(2\) (the core lies in \(H\)) and divides \(2! = 2\), so \(H\) equals its core: every subgroup of index \(2\) is normal.
  • More generally, in a finite group, a subgroup whose index is the smallest prime \(p\) dividing \(|G|\) is normal: the index of the normal core divides both \(|G|\) and \(p!\); since no prime smaller than \(p\) divides \(|G|\) and \(p^2\) does not divide \(p!\), it divides \(p\). It is also a multiple of \(p\), because the core lies in \(H\), so it equals \(p\) and \(H\) equals its core.
  • If \(G\) is simple and has a subgroup of finite index \(n > 1\), the normal core is trivial, so \(G\) embeds in \(S_n\) and \(|G|\) divides \(n!\).

Combining the left regular representation with the signature of a permutation gives another classical consequence: a group of order \(2m\) with \(m\) odd has a normal subgroup of index \(2\). An element of order \(2\), which exists by Cauchy's theorem, has no fixed points under left multiplication, so it permutes \(G\) as a product of \(m\) disjoint transpositions, an odd permutation. The elements of \(G\) that act by even permutations then form a normal subgroup of index \(2\).

In Mathlib

Mathlib proves the theorem in the generality of the second paragraph of the proof: for any faithful action of a group \(G\) on a type \(X\), it gives an isomorphism between \(G\) and the image of the homomorphism from \(G\) to the permutations of \(X\). The classical statement is the case where \(X\) is \(G\) acting on itself by left multiplication; Mathlib finds the faithfulness of that action automatically. The isomorphism is constructed with the axiom of choice and is not computable. In the smallest-prime criterion, Mathlib's cardinality assigns \(0\) to an infinite group, and the smallest prime factor of \(0\) is taken to be \(2\), so for an infinite group the statement reduces to the index-\(2\) case and needs no finiteness hypothesis.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibEquiv.Perm.subgroupOfMulAction
Uses 5
Statement usesGroup · Subgroup · Permutation group of a type · Group morphism
Proof usesGroup action

DefinitionAlternating group#

The alternating group \(A_n\) is the group of even permutations of \(n\) letters, a subgroup of index two in the symmetric group \(S_n\) when \(n \ge 2\). The alternating groups \(A_n\) with \(n \ge 5\) form the most elementary infinite family of nonabelian simple groups: each has no normal subgroups other than the trivial group and itself. This fact is the group-theoretic reason why the general polynomial equation of degree five or more cannot be solved by radicals, and the alternating groups form one of the infinite families in the classification of finite simple groups.

Definition

A transposition is a permutation that exchanges two letters and fixes the rest. Every permutation of a finite set can be written as a product of transpositions, and although such a product is not unique, the parity of the number of transpositions is. The sign of a permutation is \(+1\) if this number is even and \(-1\) if it is odd, and it defines a group morphism from the permutation group \(S_n\) to the two-element group \(\{+1, -1\}\).

The alternating group on a finite set \(X\) is the kernel of the sign morphism, so that a permutation belongs to it exactly when its sign is \(+1\). Equivalently, a product of transpositions lies in the alternating group if and only if the number of factors is even. The group \(A_n\) is the alternating group on \(\{1, \dots, n\}\); a bijection between two finite sets induces an isomorphism between their alternating groups, so only the number of letters matters.

Examples
  • For \(n \le 1\) the symmetric group is trivial, and so is \(A_n\). For \(n = 2\) the only nontrivial permutation is a transposition, which is odd, so \(A_2\) is again trivial.
  • \(A_3\) consists of the identity and the two 3-cycles, the permutations that send one letter \(a\) to \(b\), \(b\) to \(c\) and \(c\) back to \(a\); it is cyclic of order \(3\).
  • \(A_4\) has order \(12\). It contains the identity, eight 3-cycles and the three products of two disjoint transpositions; those three products together with the identity form a copy of the Klein four-group \(\mathbb{Z}/2 \times \mathbb{Z}/2\), which is a normal subgroup of \(A_4\). The group \(A_4\) is a standard example showing that the converse of Lagrange's theorem fails: it has no subgroup of order \(6\).
  • \(A_5\) has order \(60\) and is the smallest nonabelian simple group. It is isomorphic to the group of rotations of a regular icosahedron.
Properties

Since \(S_n\) is not solvable for \(n \ge 5\) (its normal subgroup \(A_n\) is simple and nonabelian), Galois theory shows that a polynomial of degree \(n \ge 5\) over a field of characteristic zero whose Galois group is the full symmetric group \(S_n\) cannot be solved by radicals (see solvability by radicals implies a solvable Galois group); applying this to a suitable quintic over the rationals gives the Abel–Ruffini theorem. Through Cayley's theorem every finite group embeds in some symmetric group, and composing with the embedding of \(S_n\) into \(A_{n+2}\) shows that every finite group also embeds in an alternating group. The quotient \(S_n / A_n\) is the quotient group that the first isomorphism theorem identifies, for \(n \ge 2\), with the image \(\{+1, -1\}\) of the sign morphism.

In Mathlib

Mathlib defines the alternating group for any finite type with decidable equality, as a subgroup of the full permutation group of that type, namely the kernel of the sign morphism; the textbook \(A_n\) is the case of a type with exactly \(n\) elements. Statements that depend on \(n\) are phrased through the number of elements of the type. The formulas for the index and the order assume at least two elements: with fewer, the index is \(1\), and the order formula would fail, since \(1/2\) rounds down to \(0\) in the division of natural numbers. Mathlib proves simplicity for \(n \ge 5\) through Iwasawa's criterion, applied to the action on three-element subsets, with the action on four-element subsets covering the case of six letters. Its statement that \(A_n\) is the commutator subgroup of \(S_n\) carries the hypothesis \(n \ge 5\), while the inclusion of the commutator subgroup in \(A_n\) is proved for every \(n\).

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibalternatingGroup
Uses 4 · Used by 1
Statement usesPermutation group of a type · Signature · Subgroup · Group morphism
Used byAlternating groups on 5 letters or more are simple

TheoremAlternating groups on 5 letters or more are simple#

The simplicity of the alternating groups is the theorem that for \(n \ge 5\) the alternating group \(A_n\), the group of even permutations of \(n\) letters, has no normal subgroups other than the trivial subgroup and itself. It gives an infinite family of simple groups that are not abelian, it is the group-theoretic reason why the general equation of degree five or more cannot be solved by radicals, and the alternating groups form one of the families in the classification of finite simple groups.

Statement

Let \(X\) be a finite set with \(n \ge 5\) elements, and let \(A_n\) be the subgroup of the permutation group \(S_n\) of \(X\) consisting of the permutations of signature \(+1\). Then \(A_n\) is a simple group: it is nontrivial, and every normal subgroup of \(A_n\) is either \(\{e\}\) or \(A_n\). Its order is \(n!/2\).

Small cases

The bound \(n \ge 5\) marks exactly where \(A_n\) becomes simple and non-abelian. For \(n \le 3\) the group \(A_n\) is cyclic (see cyclic group): it is trivial for \(n \le 2\) and of order \(3\) for \(n = 3\), so \(A_3\) is simple but abelian. For \(n = 4\) the three products of two disjoint transpositions together with the identity form a subgroup of order \(4\), the Klein four-group, and this subgroup is normal in \(A_4\), so \(A_4\) is not simple. The group \(A_5\), of order \(60\), is the smallest simple group that is not abelian; it is isomorphic to the group of rotations of a regular icosahedron.

Proof

A classical argument runs in four steps, with permutations composed from right to left.

  1. The 3-cycles generate \(A_n\). Every even permutation is a product of an even number of transpositions, and products of two transpositions are products of 3-cycles: $\((a\,b)(b\,c) = (a\,b\,c), \qquad (a\,b)(c\,d) = (a\,b\,c)(b\,c\,d).\)$ So the subgroup generated by the 3-cycles is the whole alternating group.
  2. For \(n \ge 5\) all 3-cycles are conjugate in \(A_n\). Two 3-cycles are conjugate in \(S_n\) by some permutation \(\pi\). If \(\pi\) is odd, replace it by \(\pi\,(d\,e)\), where \(d\) and \(e\) are two letters outside the support of the first 3-cycle; such letters exist because \(n \ge 5\), and the transposition \((d\,e)\) commutes with that 3-cycle. Hence any two 3-cycles are conjugate by an element of \(A_n\).
  3. A nontrivial normal subgroup contains a 3-cycle. Let \(N\) be a normal subgroup of \(A_n\) and \(\sigma \neq e\) an element of \(N\). For \(n \ge 4\) the center of \(A_n\) is trivial (the center is the subgroup of elements that commute with every element), and the 3-cycles generate \(A_n\), so some 3-cycle \(\rho\) does not commute with \(\sigma\). Write \([x, y] = x y x^{-1} y^{-1}\) for the commutator of \(x\) and \(y\). The element $\(\tau = [\sigma, \rho] = \sigma\rho\sigma^{-1}\rho^{-1} = (\sigma\rho\sigma^{-1})\,\rho^{-1}\)$ lies in \(N\), because \(N\) contains \(\sigma\) and the conjugate \(\rho\sigma^{-1}\rho^{-1}\). It is not the identity, and as a product of two 3-cycles it moves at most six letters. Such a product is a 3-cycle, a product of two disjoint transpositions, a 5-cycle, or a product of two disjoint 3-cycles. If \(\tau\) is a 3-cycle, this step is done. Otherwise the same trick, now applied to \(\tau\), yields a 3-cycle in \(N\) after one or two rounds: for a 3-cycle \(\rho'\), the commutator \([\tau, \rho']\) again lies in \(N\), and a suitable choice of \(\rho'\) makes it simpler. If \(\tau = (a\,b)(c\,d)\), its commutator with \((a\,b\,e)\), where \(e\) is a fifth letter, is the 3-cycle \((a\,b\,e)\); this is where \(n \ge 5\) enters. If \(\tau = (a\,b\,c\,d\,e)\), its commutator with \((a\,b\,c)\) is the 3-cycle \((a\,d\,b)\). If \(\tau = (a\,b\,c)(d\,e\,f)\), its commutator with \((a\,b\,d)\) is the 5-cycle \((a\,d\,c\,e\,b)\), which reduces to the previous case.
  4. Conclusion. By step 2, \(N\) contains every 3-cycle, and by step 1, \(N = A_n\).
Consequences
In Mathlib

Mathlib defines the alternating group on any finite type as the kernel of the signature homomorphism, a subgroup of the full permutation group, and the theorem is stated for every finite type with at least five elements rather than for the sets \(\{1, \dots, n\}\) only. The proof does not follow the classical case analysis. It uses Iwasawa's criterion. Suppose a group acts quasiprimitively on a set, meaning that the action is transitive and so is the action of every normal subgroup that moves some point. Primitive actions, the transitive actions that preserve no partition of the set other than the two trivial ones, qualify, since the orbits of a normal subgroup form such a partition. Suppose also that each point \(x\) is assigned an abelian subgroup \(T_x\) so that \(g T_x g^{-1} = T_{gx}\) for every element \(g\), and that the subgroups \(T_x\) together generate the group. Then every normal subgroup that moves some point contains the commutator subgroup. Mathlib applies this to the primitive action of \(A_n\) on the three-element subsets of the letters when \(n \neq 6\), with \(T_x\) the alternating group of the three letters in \(x\), and on the four-element subsets when \(n \neq 8\), with \(T_x\) the Klein four-group of the four letters in \(x\), so every \(n \ge 5\) is covered. A nontrivial normal subgroup moves some subset, because the action is faithful, so it contains the commutator subgroup, which is all of \(A_n\) by perfectness. Mathlib also packages the criterion as a simplicity test for a nontrivial perfect group acting faithfully, but the proof for \(A_n\) goes through the commutator form. The non-solvability of the full permutation group is proved for any type of cardinality at least five, finite or infinite, by an explicit computation in \(S_5\) rather than from simplicity.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibalternatingGroup.isSimpleGroup
Uses 9
Statement usesAlternating group · Simple group · Normal subgroup · Permutation group of a type
Proof usesSubgroup generated by a subset · Signature · Decomposition into transpositions · Conjugacy classes · Decomposition into cycles with disjoint support

TheoremClassification of abelian simple groups#

The classification of abelian simple groups says that an abelian group is simple exactly when its order is a prime number, so that up to isomorphism the abelian simple groups are the cyclic groups \(\mathbb{Z}/p\mathbb{Z}\), one for each prime \(p\). Finite simple groups are the building blocks from which every finite group is assembled by successive extensions, as the factors of a composition series, and this theorem settles the commutative case completely: every other simple group is nonabelian, which is where the difficulty of the classification of finite simple groups lies.

Statement

A group \(G\) is simple if it is nontrivial and its only normal subgroups are the trivial subgroup and \(G\) itself (the definition). The theorem reads:

Let \(G\) be an abelian group. Then \(G\) is simple if and only if its order is a prime number.

In particular an abelian simple group is automatically finite, and it is cyclic, generated by any element other than the identity.

Examples
  • For each prime \(p\), the integers modulo \(p\) under addition form a simple group: the additive group of \(\mathbb{Z}/p\mathbb{Z}\) is simple.
  • The group \(\mathbb{Z}/6\mathbb{Z}\) is abelian but not simple, since the multiples of \(2\) form a subgroup of order \(3\), which is normal because the group is abelian.
  • The integers under addition are not simple: the even integers form a proper nontrivial subgroup. Likewise the rationals contain the integers as a proper nontrivial subgroup. No infinite abelian group is simple.
  • Simple groups that are not abelian exist, for instance the alternating groups \(A_n\), which are simple for \(n \ge 5\). The group \(A_5\), of order \(60\), is the smallest nonabelian simple group.
Proof

One direction needs no commutativity at all. If \(|G| = p\) is prime, then by Lagrange's theorem the order of every subgroup divides \(p\), so it is \(1\) or \(p\), and every subgroup is either trivial or all of \(G\). Since \(p > 1\) the group is nontrivial, hence every group of prime order is simple.

For the converse, let \(G\) be abelian and simple. In an abelian group every subgroup is normal, so simplicity says that \(G\) has no subgroups other than \(\{e\}\) and \(G\). Choose an element \(g \neq e\). The cyclic subgroup it generates is not trivial, so it is all of \(G\), and \(G\) is cyclic. If \(g\) had infinite order, the subgroup generated by \(g^2\) would be proper and nontrivial, which is impossible; so \(G\) is finite of some order \(n > 1\). If \(n\) factored as \(n = ab\) with \(1 < a < n\), the subgroup generated by \(g^a\) would have order \(b\), strictly between \(1\) and \(n\), again impossible. Hence \(n\) is prime: an abelian simple group has prime order, and in particular it is finite.

Consequences
  • Groups of prime order. Combining the two directions, a finite group of prime order is simple and cyclic, hence abelian, so the groups of prime order are precisely the abelian simple groups.
  • Composition factors. By the Jordan–Hölder theorem every finite group has simple composition factors that are determined by the group up to isomorphism and order. The theorem identifies which of them can be abelian: exactly the cyclic groups of prime order. A finite group is solvable precisely when all of its composition factors are of this kind.
  • Odd order. The Feit–Thompson theorem says that every finite group of odd order is solvable. Together with this classification, it implies that a finite simple group of odd order is cyclic of prime order.

The cyclic groups of prime order, which this theorem identifies, form the abelian, and easiest, family in the classification of finite simple groups, whose other members (the alternating groups, the groups of Lie type and the sporadic groups) are all nonabelian. More generally, the structure theorem for finitely generated abelian groups describes every finitely generated abelian group, not only the simple ones, as a product of cyclic groups. The abelianization of a group is its largest abelian quotient; a nonabelian simple group has trivial abelianization.

In Mathlib

Mathlib measures the order of a group with a cardinality that is \(0\) for an infinite set. Since \(0\) is not prime, the statement that an abelian group is simple exactly when its order is prime already contains the fact that no infinite abelian group is simple, and Mathlib records that consequence separately. Simplicity is a property of a group rather than extra structure, and the definition includes nontriviality, so the trivial group is not simple. The theorem is stated for groups with a bundled commutative structure; a companion version asks only that the multiplication of an ordinary group be commutative (the same equivalence), and the direction from prime order to simplicity holds for arbitrary groups. The additive form, for groups written with \(+\), is stated as well.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibCommGroup.is_simple_iff_prime_card
Uses 8
Statement usesGroup · Simple group · Normal subgroup · Prime numbers
Proof usesLagrange's theorem · Cyclic group · Order of an element · Subgroup generated by a subset

TheoremStructure of finitely generated abelian groups#

The structure theorem for finitely generated abelian groups says that every abelian group generated by finitely many elements is a direct product of finitely many cyclic groups: some copies of the integers and some finite cyclic groups of prime-power order. It classifies these groups completely up to isomorphism. Questions about such a group, such as how many elements of a given order it has or which groups occur as its subgroups, reduce to questions about a list of integers, and the theorem is the model case of the structure theorem for finitely generated modules over a principal ideal domain.

Statement

An abelian group \(G\), written additively, is finitely generated if it contains finitely many elements \(g_1, \dots, g_k\) such that every element of \(G\) can be written as \(a_1 g_1 + \dots + a_k g_k\) with integers \(a_i\). For such a group there are a natural number \(r\), primes \(p_1, \dots, p_m\) (not necessarily distinct) and positive exponents \(e_1, \dots, e_m\) such that

\[G \cong \mathbb{Z}^r \times \mathbb{Z}/p_1^{e_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_m^{e_m}\mathbb{Z}.\]

Mathlib states the same decomposition with the exponents allowed to be any natural numbers, so that some factors may be trivial groups \(\mathbb{Z}/p^0\mathbb{Z}\); dropping those factors gives the display above with positive exponents. The number \(r\) is the rank of \(G\), and the prime powers \(p_i^{e_i}\) are its elementary divisors. Both are determined by \(G\) (the elementary divisors counted with multiplicity and up to order), so two finitely generated abelian groups are isomorphic exactly when they have the same rank and the same elementary divisors.

Grouping the elementary divisors differently gives an equivalent invariant factor form: \(G \cong \mathbb{Z}^r \times \mathbb{Z}/d_1\mathbb{Z} \times \cdots \times \mathbb{Z}/d_s\mathbb{Z}\) with integers \(d_1, \dots, d_s > 1\) and each \(d_i\) dividing \(d_{i+1}\). The two forms are related by the Chinese remainder theorem: \(\mathbb{Z}/mn\mathbb{Z} \cong \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}\) when \(m\) and \(n\) are coprime.

Examples
  • The cyclic group \(\mathbb{Z}/12\mathbb{Z}\) has elementary divisors \(4\) and \(3\), so it is isomorphic to \(\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}\). It is not isomorphic to \(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}\), whose elementary divisors are \(2\), \(2\) and \(3\).
  • Up to isomorphism there are exactly three abelian groups of order \(8\), one for each partition of \(3\) (each way of writing \(3\) as a sum of positive integers, ignoring order): \(\mathbb{Z}/8\mathbb{Z}\), \(\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\) and \((\mathbb{Z}/2\mathbb{Z})^3\). In general the number of abelian groups, up to isomorphism, of order \(p_1^{a_1} \cdots p_t^{a_t}\) with distinct primes \(p_i\) is the product of the numbers of partitions of the exponents \(a_1, \dots, a_t\).
  • The additive group of the rationals is abelian but not finitely generated, and it is not a product of cyclic groups: any two nonzero rationals have a common nonzero multiple, so it has no nontrivial decomposition at all, and it is not cyclic.
Consequences
  • Finite abelian groups. A finite abelian group has rank \(0\), so it is a direct sum of cyclic groups of prime-power order (for finitely many factors the direct sum and the direct product are the same group), with trivial factors allowed, and, discarding those, a direct sum of cyclic groups each of order greater than \(1\).
  • Torsion. In a finitely generated abelian group the elements of finite order form a subgroup \(T\), the torsion subgroup, isomorphic to the finite part of the decomposition. In particular, a finitely generated abelian group in which every element has finite order is finite.
  • Torsion-free groups. A finitely generated abelian group with no nonzero elements of finite order is isomorphic to \(\mathbb{Z}^r\) for some \(r\); more generally, over a principal ideal domain, a finitely generated module is free exactly when it is torsion-free (that is, \(a m = 0\) implies \(a = 0\) or \(m = 0\)).
  • A converse to Lagrange's theorem. For a finite abelian group, every divisor of the order is the order of some subgroup. For arbitrary finite groups this converse of Lagrange's theorem fails: the alternating group \(A_4\) of order \(12\) has no subgroup of order \(6\). In the abelian case, write the divisor \(d\) as a product \(d_1 \cdots d_m\) with each \(d_i\) dividing the order \(n_i\) of the \(i\)-th cyclic factor (for instance, take \(d_1 = \gcd(d, n_1)\) and repeat with \(d/d_1\) and the remaining factors); a cyclic group has a subgroup of every order dividing its own, and the product of the subgroups of order \(d_i\) of these factors has order \(d\).
  • Cyclicity. A finite abelian group whose order is not divisible by the square of any prime is cyclic, since its elementary divisors are distinct primes and the Chinese remainder theorem combines them.
Proof

Choose generators \(g_1, \dots, g_k\) of \(G\). The map \(\mathbb{Z}^k \to G\) sending \((a_1, \dots, a_k)\) to \(a_1 g_1 + \dots + a_k g_k\) is a surjective homomorphism, so \(G \cong \mathbb{Z}^k / K\) for its kernel \(K\). Because \(\mathbb{Z}\) is a principal ideal domain, a subgroup of \(\mathbb{Z}^k\) is again free of finite rank. More precisely, after renumbering, there are a basis \(f_1, \dots, f_k\) of \(\mathbb{Z}^k\) and integers \(d_1, \dots, d_s\) such that \(d_1 f_1, \dots, d_s f_s\) is a basis of \(K\); the \(d_i\) are nonzero, since the vectors \(d_i f_i\) form a basis. In matrix language this is the Smith normal form of an integer matrix whose columns generate \(K\), where one can also arrange that the \(d_i\) are positive and each divides the next; Mathlib's version does not record that divisibility, and the proof does not need it. Replacing \(d_i\) by \(|d_i|\), in these coordinates

\[G \cong \mathbb{Z}/d_1\mathbb{Z} \times \cdots \times \mathbb{Z}/d_s\mathbb{Z} \times \mathbb{Z}^{k-s}.\]

Factoring each \(d_i\) into prime powers, using the fundamental theorem of arithmetic, and applying the Chinese remainder theorem splits each finite factor into cyclic groups of prime-power order. Uniqueness follows by counting: the rank is the rank of the free abelian group \(G/T\), where \(T\) is the torsion subgroup, and for each prime \(p\) and each \(t \ge 1\) the number of elementary divisors equal to \(p^j\) with \(j \ge t\) is the dimension over \(\mathbb{Z}/p\mathbb{Z}\) of the subgroup of elements of \(p^{t-1} G = \{p^{t-1} g : g \in G\}\) that are killed by \(p\).

Generalizations

An abelian group is the same thing as a module over the integers, and being finitely generated as a group is the same as being finitely generated as a module over the integers. The theorem is therefore the special case \(R = \mathbb{Z}\) of the statement that every finitely generated module over a principal ideal domain \(R\) is a free module of finite rank times a direct sum of quotients \(R/(p^e)\) with \(p\) irreducible. The case of the polynomial ring over a field yields the rational canonical form of a matrix and, when the characteristic polynomial splits into linear factors, its Jordan canonical form.

In Mathlib

Mathlib proves the theorem by applying its structure theorem for modules over a principal ideal domain to the integers. Rather than through Smith normal form, the module theorem is proved by splitting off the torsion submodule, whose quotient is torsion-free and hence free, and decomposing the torsion part first into its primary components and then into cyclic pieces. The free part \(\mathbb{Z}^r\) appears as the group of integer-valued functions on a set of \(r\) elements, and the finite part as a direct sum indexed by a finite type. The exponents are only required to be natural numbers, so the statement allows trivial factors \(\mathbb{Z}/p^0\mathbb{Z}\); a separate version for finite groups removes them, at the cost of stating only that each cyclic factor has order greater than \(1\). Only the existence of the decomposition is stated, not the uniqueness of the rank and elementary divisors. A multiplicative version states the same result for abelian groups written multiplicatively, with the direct sum replaced by a product.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibAddCommGroup.equiv_free_prod_directSum_zmod
Uses 14
Statement usesGroup · Subgroup generated by a subset · Cyclic group · Direct product of groups · Prime numbers · Quotient group · Group morphism · Integer
Proof usesPrincipal ideal domain · Chinese remainder theorem · The Fundamental Theorem of Arithmetic · First isomorphism theorem · Basis · Order of an element

TheoremLagrange's theorem#

Lagrange's theorem says that in a finite group the order (number of elements) of every subgroup divides the order of the group. It is a basic structural result of group theory: it places an arithmetic restriction on the possible orders of subgroups, and several classical facts, Fermat's little theorem among them, follow from it in a line.

Statement

Let \(G\) be a finite group and \(H\) a subgroup of \(G\), and write \(|G|\) for the order of \(G\). Then the order \(|H|\) of \(H\) divides the order \(|G|\) of \(G\). More precisely,

\[|G| = [G : H]\,|H|,\]

where the index \([G : H]\) is the number of left cosets of \(H\) in \(G\) (the counting formula). In particular the index also divides the order of \(G\).

Proof

The left cosets \(gH\) partition \(G\): two cosets are either equal or disjoint, and every element \(g\) lies in its own coset \(gH\). Each coset has exactly \(|H|\) elements, because \(h \mapsto gh\) is a bijection from \(H\) onto \(gH\). Counting the elements of \(G\) coset by coset gives \(|G| = [G : H]\,|H|\).

Consequences
The converse fails

A divisor of \(|G|\) need not be the order of any subgroup: the alternating group \(A_4\) has order \(12\) but no subgroup of order \(6\). Two partial converses hold.

In Mathlib

Mathlib states the theorem for every group, not only finite ones. It measures size with the natural-number cardinality, which gives \(0\) for an infinite set, so for an infinite group the statement holds automatically, since every number divides \(0\); for a finite group it is the classical theorem. The index is likewise the number of cosets as a natural number, so it is \(0\) exactly when \(H\) has infinitely many cosets, and an element of infinite order has order \(0\). With these conventions the counting formula, the divisibility of the order by the index and by element orders, and the statement about chains of subgroups hold for every group. The statement about groups of prime order needs no convention, since its hypothesis already makes the group finite. Mathlib's statements of Cauchy's theorem and of the first Sylow theorem assume the group is finite.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibSubgroup.card_subgroup_dvd_card
Uses 3 · Used by 6
Statement usesGroup · Subgroup · Index of a subgroup
Used byClass formula · Classification of abelian simple groups · First Sylow theorem · Second Sylow theorem · Third Sylow theorem · General linear group (Classical automorphism groups)

TheoremFirst Sylow theorem#

The first Sylow theorem says that a finite group has a subgroup of every prime-power order that divides the order of the group. It is a partial converse to Lagrange's theorem, which says that the order of a subgroup divides the order of the group. A general divisor of \(|G|\) need not be the order of any subgroup, but a divisor that is a power of a single prime always is. Together with the second and third Sylow theorems, it is the basic tool for analysing finite groups of a given order.

Statement

Let \(G\) be a finite group and \(p\) a prime. If \(p^n\) divides \(|G|\), then \(G\) has a subgroup of order \(p^n\).

Write \(|G| = p^k m\) with \(p \nmid m\). The case \(n = k\) is the usual form of the theorem: \(G\) has a subgroup of order \(p^k\), the largest power of \(p\) dividing \(|G|\). Such a subgroup is called a Sylow \(p\)-subgroup of \(G\). A subgroup whose order is a power of \(p\) is a \(p\)-subgroup.

The case \(n = 1\) is Cauchy's theorem: if \(p\) divides \(|G|\), then \(G\) has an element of order \(p\), whose powers form a subgroup of order \(p\).

Examples
  • The symmetric group \(S_3\) has order \(6 = 2 \cdot 3\). Each of its three transpositions generates a Sylow \(2\)-subgroup of order \(2\), and the alternating group \(A_3\), of order \(3\), is its Sylow \(3\)-subgroup.
  • The alternating group \(A_4\) has order \(12 = 2^2 \cdot 3\). Its subgroups of order \(2\), \(3\) and \(4\) exist as the theorem predicts; the subgroup of order \(4\) is the Klein four-group, consisting of the identity and the three double transpositions. The divisor \(6\) is not a prime power, and \(A_4\) has no subgroup of order \(6\).
  • In the general linear group of invertible \(d \times d\) matrices over the field with \(p\) elements, the upper triangular matrices with ones on the diagonal form a subgroup of order \(p^{d(d-1)/2}\). This is the full power of \(p\) in the order $\((p^d - 1)(p^d - p)\cdots(p^d - p^{d-1}),\)$ so these matrices form a Sylow \(p\)-subgroup.
Consequences
  • Towers of subgroups. If \(p^n\) divides \(|G|\), there is a chain of subgroups \(H_0 \le H_1 \le \dots \le H_n\) with \(|H_i| = p^i\) for each \(i\) (towers of \(p\)-subgroups). In particular a group of order \(p^n\) has subgroups of every order \(p^i\) with \(i \le n\).
  • Extending a \(p\)-subgroup. A subgroup of order \(p^m\) is contained in a subgroup of order \(p^n\) whenever \(m \le n\) and \(p^n\) divides \(|G|\) (the extension statement). Taking \(n = k\) shows that every \(p\)-subgroup lies in some Sylow \(p\)-subgroup.
  • Maximal \(p\)-subgroups. In a finite group, a \(p\)-subgroup \(P\) contained in no larger \(p\)-subgroup has order \(p^k\), the full power of \(p\) in \(|G|\), so it is a Sylow \(p\)-subgroup in the sense above. (Mathlib takes maximality among \(p\)-subgroups as the definition; see In Mathlib.) Moreover, its index \([G : P]\) is not divisible by \(p\); Mathlib proves the index statement for any Sylow subgroup of finite index, in any group.
  • The other Sylow theorems. Once Sylow subgroups exist, one can ask how many there are and how they are related: they are all conjugate, and their number divides \(m\) and is congruent to \(1\) modulo \(p\).
Proof

The theorem follows by induction on \(n\) from the following step: if \(H\) is a subgroup of order \(p^n\) and \(p^{n+1}\) divides \(|G|\), then \(H\) lies in a subgroup of order \(p^{n+1}\). Starting from the trivial subgroup, of order \(p^0 = 1\), the step is applied \(n\) times.

For the step, let \(H\) act on the set of left cosets \(G/H\) by left multiplication (see group action). Every orbit of a group of order \(p^n\) has size a power of \(p\), so only the fixed points contribute an orbit of size not divisible by \(p\); hence the number of fixed points is congruent to \(|G/H|\) modulo \(p\). Since \(|G| = p^n\,[G : H]\) and \(p^{n+1}\) divides \(|G|\), the index \([G : H]\) is divisible by \(p\), so the number of fixed points is too.

A coset \(gH\) is fixed by every element of \(H\) exactly when \(g^{-1} h g \in H\) for all \(h \in H\), that is, when \(g^{-1} H g \subseteq H\). Since \(H\) is finite, this inclusion is an equality, so \(g\) lies in the normalizer \(N(H) = \{x \in G : xHx^{-1} = H\}\), a subgroup of \(G\) that contains \(H\) as a normal subgroup (indeed the largest such subgroup). The fixed cosets are therefore the cosets of \(H\) in \(N(H)\), and \(p\) divides the order of the quotient group \(N(H)/H\). By Cauchy's theorem this quotient has a subgroup of order \(p\). Its preimage in \(N(H)\) is a subgroup containing \(H\), of order \(p \cdot p^n = p^{n+1}\).

A different proof counts the subsets of \(G\) of size \(p^k\) and lets \(G\) act on them by left multiplication; it avoids Cauchy's theorem.

In Mathlib

Mathlib states the theorem for a finite group, measuring the sizes of the group and the subgroup by their natural-number cardinalities, and supplying the primality of \(p\) as a typeclass assumption rather than as an explicit hypothesis; for a general prime the user supplies that assumption from a proof of primality. Its stronger form, that the subgroup of order \(p^n\) can be chosen to contain any given subgroup of order \(p^m\) with \(m \le n\), is the statement the proof actually establishes. Mathlib defines a Sylow \(p\)-subgroup not by its order but as a maximal \(p\)-subgroup, where a \(p\)-group is a group in which each element \(g\) satisfies \(g^{p^i} = e\) for some \(i\). This makes sense in every group, finite or not, and Zorn's lemma then shows that every subgroup that is a \(p\)-group is contained in a Sylow \(p\)-subgroup in any group. For a finite group the two definitions agree: a maximal \(p\)-subgroup has order \(p^k\), and conversely a subgroup of order \(p^k\) is a Sylow subgroup.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibSylow.exists_subgroup_card_pow_prime
Uses 10 · Used by 1
Statement usesGroup · Subgroup · Prime numbers
Proof usesLagrange's theorem · Group action · Orbit · Index of a subgroup · Normal subgroup · Quotient group · Cauchy's theorem (group theory)
Used byThird Sylow theorem

TheoremSecond Sylow theorem#

The second Sylow theorem says that in a finite group, for each prime \(p\), any two Sylow \(p\)-subgroups are conjugate. The first Sylow theorem guarantees that Sylow subgroups exist; the second says that, up to conjugation, there is only one of them. Together with the third Sylow theorem, which constrains how many there are, it is the main tool for analysing finite groups of a given order, and in particular for showing that groups of many orders cannot be simple.

Statement

Let \(G\) be a finite group and \(p\) a prime, and write \(|G| = p^n m\) with \(p \nmid m\). A Sylow \(p\)-subgroup of \(G\) is a subgroup of order \(p^n\), the largest power of \(p\) dividing \(|G|\). Equivalently, it is a subgroup in which every element has order a power of \(p\) and which is maximal with that property. For a finite group, every element has \(p\)-power order exactly when the group has \(p\)-power order (one direction uses Cauchy's theorem), so a maximal such subgroup has order \(p^k\) for some \(k \le n\). That \(k = n\) takes more work. A subgroup of order \(p^n\) exists by the first Sylow theorem, and the fixed-point argument in the proof below shows that every subgroup of \(p\)-power order lies in a conjugate of it; so a maximal one has order exactly \(p^n\). Conversely, by Lagrange's theorem no subgroup of \(p\)-power order properly contains one of order \(p^n\), so a subgroup of order \(p^n\) is a Sylow \(p\)-subgroup.

The group \(G\) acts on the set of its Sylow \(p\)-subgroups by conjugation, \(g \cdot P = gPg^{-1}\). The theorem says that this action is transitive: for any two Sylow \(p\)-subgroups \(P\) and \(Q\) there is some \(g \in G\) with

\[Q = gPg^{-1}.\]

Since conjugation by \(g\) is an automorphism of \(G\), it follows that any two Sylow \(p\)-subgroups are isomorphic.

Textbooks number the Sylow theorems differently. Many include in the second theorem the stronger assertion that every subgroup of \(p\)-power order lies in some conjugate of a given Sylow \(p\)-subgroup; this follows by combining the statement above with the fact that every such subgroup lies in some Sylow \(p\)-subgroup.

Examples
  • In the symmetric group \(S_3\), of order \(6 = 2 \cdot 3\), the Sylow \(2\)-subgroups are the three subgroups generated by the transpositions \((1\,2)\), \((1\,3)\) and \((2\,3)\). Conjugating a transposition gives another transposition, and every transposition arises this way, so the three subgroups form a single conjugacy class. The unique Sylow \(3\)-subgroup is the alternating group \(A_3\).
  • In an abelian group every subgroup is normal, so conjugation fixes each Sylow \(p\)-subgroup. The theorem then forces there to be exactly one Sylow \(p\)-subgroup for each prime \(p\).
Consequences
Proof

The proof uses one fact about group actions of \(p\)-groups: if a group of \(p\)-power order acts on a finite set \(X\), then \(|X|\) is congruent modulo \(p\) to the number of fixed points, because by orbit–stabilizer each orbit has size equal to the index of a stabilizer, which divides the order of the acting group, so every orbit that is not a single point has size a positive power of \(p\). In particular, if \(p \nmid |X|\) then the action has a fixed point.

Let \(P\) and \(Q\) be Sylow \(p\)-subgroups and let \(X\) be the orbit of \(P\) under conjugation by \(G\). A subgroup \(H\) of \(p\)-power order fixes a Sylow \(p\)-subgroup \(S\) under conjugation exactly when \(H\) normalizes \(S\), and then \(HS\) is a subgroup of \(p\)-power order containing \(S\); by maximality, \(H\) fixes \(S\) if and only if \(H \le S\).

Restrict the action to \(P\). A fixed point \(S \in X\) satisfies \(P \le S\), and since both have the same order, \(S = P\). So \(P\) has exactly one fixed point on \(X\), and \(|X| \equiv 1 \pmod p\). Now restrict the action to \(Q\). Since \(p \nmid |X|\), the subgroup \(Q\) fixes some \(S \in X\), so \(Q \le S\) and hence \(Q = S\). Thus \(Q\) lies in the orbit of \(P\), that is, \(Q = gPg^{-1}\) for some \(g\).

In Mathlib

Mathlib defines a Sylow \(p\)-subgroup for an arbitrary group, finite or not, as a maximal subgroup among those in which every element has order a power of \(p\), and it proves the existence of such subgroups in that generality by Zorn's lemma. The second Sylow theorem is stated under the weaker hypothesis that there are only finitely many Sylow \(p\)-subgroups, which holds whenever \(G\) is finite. The conclusion is phrased as the conjugation action being pretransitive: any two points lie in one orbit, with nonemptiness proved separately. Some finiteness hypothesis is needed. In the infinite dihedral group the Sylow \(2\)-subgroups are the subgroups of order \(2\) generated by reflections, and these reflections fall into two conjugacy classes. The formulas for \(n_p\) above and the uniqueness of a normal Sylow subgroup carry the same hypothesis (the identification of the stabilizer with the normalizer needs none), while Frattini's argument asks for it only in the normal subgroup \(N\).

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibSylow.isPretransitive_of_finite
Uses 8 · Used by 1
Statement usesGroup · Subgroup · Group action · Order of an element
Proof usesOrbit · Lagrange's theorem · Index of a subgroup · Stabilizer of a point
Used byThird Sylow theorem

TheoremThird Sylow theorem#

The third Sylow theorem says that in a finite group \(G\) the number of Sylow \(p\)-subgroups, for a prime \(p\), is congruent to \(1\) modulo \(p\) and divides the order of \(G\). Together with the first and second Sylow theorems it is the basic counting tool of finite group theory: the two arithmetic conditions often leave only one possible value for the number of Sylow subgroups, and when that value is \(1\) the unique Sylow \(p\)-subgroup is a normal subgroup.

Statement

Let \(G\) be a finite group of order \(p^n m\), where \(p\) is prime and \(p \nmid m\). A \(p\)-group is a group in which every element has order a power of \(p\). A Sylow \(p\)-subgroup of \(G\) is a maximal \(p\)-subgroup: a subgroup that is a \(p\)-group and is contained in no larger such subgroup. In a finite group these are exactly the subgroups of order \(p^n\); in particular every Sylow \(p\)-subgroup has order \(p^n\), and the first Sylow theorem shows that they exist. Write \(n_p\) for the number of Sylow \(p\)-subgroups. Then

\[n_p \equiv 1 \pmod p \qquad \text{and} \qquad n_p \mid m.\]

The congruence \(n_p \equiv 1 \pmod p\) (see congruence) implies at once that \(p\) does not divide \(n_p\). For any Sylow \(p\)-subgroup \(P\), with normalizer \(N_G(P)\) (the set of elements \(g\) with \(gPg^{-1} = P\)), the number of Sylow \(p\)-subgroups equals the index of \(N_G(P)\), and \(n_p\) divides the index of \(P\), which is \(m\) by Lagrange's theorem (see index).

Examples
  • In the symmetric group \(S_3\) there are three subgroups of order \(2\), and \(3 \equiv 1 \pmod 2\); there is one subgroup of order \(3\).
  • In the alternating group \(A_5\), of order \(60\), there are six subgroups of order \(5\), ten of order \(3\) and five of order \(4\), matching \(6 \equiv 1 \pmod 5\), \(10 \equiv 1 \pmod 3\) and \(5 \equiv 1 \pmod 2\).
  • For a group of order \(15\), \(n_3\) divides \(5\) and is \(1\) modulo \(3\), so \(n_3 = 1\); likewise \(n_5\) divides \(3\) and is \(1\) modulo \(5\), so \(n_5 = 1\). Both Sylow subgroups are therefore normal, and the group is the direct product of cyclic groups of orders \(3\) and \(5\), hence cyclic.
Consequences
  • Uniqueness and normality. By the second Sylow theorem all Sylow \(p\)-subgroups are conjugate, so a Sylow \(p\)-subgroup is normal exactly when it is the only one: a normal Sylow \(p\)-subgroup is unique, and a unique Sylow \(p\)-subgroup is normal.
  • Simplicity tests. If the two conditions force \(n_p = 1\) for some prime \(p\) dividing \(|G|\) with \(p^n < |G|\), then \(G\) is not simple. For instance, in a group of order \(20\), \(n_5\) divides \(4\) and is \(1\) modulo \(5\), so \(n_5 = 1\) and no group of order \(20\) is simple. Arguments of this kind, refined by counting elements, settle the simplicity question for most small orders.
Proof

Let \(P\) be a Sylow \(p\)-subgroup and let \(P\) act on the set of all Sylow \(p\)-subgroups by conjugation, a group action. When a \(p\)-group acts on a finite set, every orbit has size a power of \(p\), by the orbit–stabilizer relation, so the orbits of size greater than \(1\) contribute multiples of \(p\), and the size of the set is congruent modulo \(p\) to the number of fixed points. It remains to show that \(P\) is the only fixed point.

A Sylow subgroup \(Q\) is fixed when \(P\) normalizes \(Q\). In that case \(PQ\) is a subgroup in which \(Q\) is normal, and its order \(|P|\,|Q| / |P \cap Q|\) is a power of \(p\). Maximality of \(Q\) gives \(PQ = Q\), so \(P \le Q\); in other words, \(Q\) is fixed by \(P\) exactly when \(P \le Q\). Since \(P\) and \(Q\) have the same order, \(Q = P\). Hence \(n_p \equiv 1 \pmod p\).

For the divisibility, let all of \(G\) act by conjugation instead. By the second Sylow theorem this action has a single orbit, and the stabilizer of \(P\) is its normalizer, so the orbit–stabilizer theorem gives \(n_p = [G : N_G(P)]\). Since \(P \le N_G(P)\), this index divides \([G : P] = m\).

In Mathlib

Mathlib works with an arbitrary group, finite or not. A Sylow \(p\)-subgroup is defined as a maximal \(p\)-subgroup, where a \(p\)-group is one in which every element has order a power of \(p\), and the Sylow subgroups form a type of their own on which the group acts by conjugation. The theorem assumes only that \(p\) is prime and that there are finitely many Sylow \(p\)-subgroups, so it covers some infinite groups as well; for a finite group this hypothesis holds automatically and the statement is the classical one. The number of Sylow subgroups is the natural-number cardinality of that type, which is finite by hypothesis; Mathlib's natural-number cardinality would assign \(0\) to an infinite type, so the finiteness assumption is what makes the count meaningful. The second Sylow theorem and the identification of \(n_p\) with the index of a normalizer are proved under the same finiteness assumption.

Written for Trough, under CC BY-SA 4.0.

Sources
Mathlibcard_sylow_modEq_one
Uses 12 · Used by 2
Statement usesGroup · Subgroup · Prime numbers · Congruence in \(\Z\)
Proof usesFirst Sylow theorem · Group action · Orbit · Stabilizer of a point · Second Sylow theorem · Normal subgroup · Lagrange's theorem · Index of a subgroup
Used byFocal subgroup theorem · Schur–Zassenhaus theorem

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