1 Introduction
If H is a subgroup of a finite group G, what can we say about the relationship between ∣H∣ and ∣G∣? Consider, for instance, S3, which has order 6. Its subgroups have orders 1,2,3, and 6, each of which divides 6. This is no coincidence.
Lagrange's theorem establishes this divisibility in full generality. The key idea is the notion of a coset.
2 Definition of Cosets
Let G be a group, H≤G, and a∈G. The sets aHHa={ah∣h∈H}(the left coset of H containing a)={ha∣h∈H}(the right coset of H containing a)
are called the left and right cosets of H in G determined by a.
Let G=S3 and H={e,(12)}, a subgroup of order 2. The left cosets of H are: eH=(12)H(13)H=(123)H(23)H=(132)H={e,(12)}={(13),(123)}={(23),(132)}
These three cosets partition S3.
Let G=(Z,+) and H=3Z. The cosets (written additively as a+H) are: 0+3Z1+3Z2+3Z={…,−6,−3,0,3,6,…}={…,−5,−2,1,4,7,…}={…,−4,−1,2,5,8,…}
This is simply the classification of integers by their remainder upon division by 3— the origin of the term "residue class."
3 Basic Properties of Cosets
Let H≤G. The left cosets of H in G satisfy the following:
Every element belongs to some coset: a∈aH for all a∈G.
aH=bH if and only if a−1b∈H.
Distinct cosets are disjoint: aH=bH implies aH∩bH=∅.
All left cosets have the same cardinality: ∣aH∣=∣H∣.
Consequently, the left cosets of H form a partition of G.
(1) a=ae∈aH.(2) If aH=bH, then b=be∈bH=aH, so b=ah for some h∈H. Hence a−1b=h∈H. Conversely, if a−1b=h∈H, then b=ah, and for any bh′∈bH we have bh′=ahh′∈aH (since hh′∈H), giving bH⊆aH. Similarly, a=bh−1 implies aH⊆bH.(3) Suppose aH∩bH=∅, and pick c∈aH∩bH. Then c=ah1=bh2 for some h1,h2∈H, so a−1b=h1h2−1∈H. By part (2), aH=bH. Taking the contrapositive gives (3).(4) The map f:H→aH defined by f(h)=ah is a bijection (surjectivity is by definition; injectivity follows from the cancellation law). □
For H≤G, define a relation ∼ on G by a∼b⟺a−1b∈H.
Then ∼ is an equivalence relation, and the equivalence class of a is aH.
Reflexivity: a−1a=e∈H. Symmetry: a−1b∈H⇒(a−1b)−1=b−1a∈H. Transitivity: a−1b,b−1c∈H⇒a−1c=(a−1b)(b−1c)∈H. □
4 Lagrange's Theorem
For H≤G, the number of left cosets of H in G is called the index of H in G and is denoted [G:H].
Let G be a finite group and H≤G. Then ∣G∣=[G:H]⋅∣H∣.
In particular, ∣H∣ divides ∣G∣.
Let the distinct left cosets of H be a1H,a2H,…,akH, where k=[G:H]. These cosets partition G: G=a1H⊔a2H⊔⋯⊔akH(disjoint union).
Since each coset has cardinality ∣H∣, ∣G∣=k⋅∣H∣=[G:H]⋅∣H∣.
□
5 Corollaries of Lagrange's Theorem
Lagrange's theorem has numerous important consequences.
For an element a of a finite group G, ord(a) divides ∣G∣. In particular, a∣G∣=e.
We have ord(a)=∣⟨a⟩∣ and ⟨a⟩≤G, so Lagrange's theorem gives ord(a)∣∣G∣. Writing ∣G∣=ord(a)⋅m, we find a∣G∣=(aord(a))m=em=e. □
If ∣G∣=p where p is prime, then G is cyclic and G≅Z/pZ.
Pick any a∈G with a=e. Since ord(a) divides ∣G∣=p and ord(a)=1 (because a=e), we must have ord(a)=p. Therefore ⟨a⟩=G. □
Let p be a prime and let a be an integer with gcd(a,p)=1. Then ap−1≡1(modp).
The set (Z/pZ)×={1ˉ,2ˉ,…,p−1} is a group of order p−1 under multiplication (since p is prime, every nonzero element has a multiplicative inverse). Since gcd(a,p)=1, we have aˉ∈(Z/pZ)×. By the corollary to Lagrange's theorem, aˉp−1=1ˉ, i.e. ap−1≡1(modp). □
Let n be a positive integer and let a be an integer with gcd(a,n)=1. Then aφ(n)≡1(modn), where φ(n) denotes Euler's totient function (the number of integers between 1 and n coprime to n).
The set (Z/nZ)×={aˉ∣gcd(a,n)=1} is a group of order φ(n) under multiplication. By the corollary to Lagrange's theorem, aˉφ(n)=1ˉ. □
6 The Converse of Lagrange's Theorem Fails
If d∣∣G∣, the group G need not possess a subgroup of order d.
Consider the alternating group A4 of order 12. We have 6∣12, yet A4 has no subgroup of order 6.To see this, we enumerate the elements of A4:
The identity e (1 element).
The 3-cycles (abc) (8 elements): (123),(132),(124),(142),(134),(143),(234),(243).
Products of two disjoint transpositions (ab)(cd) (3 elements): (12)(34),(13)(24),(14)(23).
The total is 12=4!/2.Suppose a subgroup H of order 6 exists. Since [A4:H]=2, the subgroup H is normal in A4 (a subgroup of index 2 is always normal, as we shall prove in the next chapter).Since H is normal, it must be a union of conjugacy classes of A4. The conjugacy classes of A4 are: {e}(1),{(12)(34),(13)(24),(14)(23)}(3),{(123),(124),(134),(234)}(4),{(132),(142),(143),(243)}(4).
The sizes are 1,3,4,4. To get ∣H∣=6, we would need a sub-sum of 1+3+4+4 equalling 6. But 1+3=4, 1+4=5, 1+3+4=8— none give 6. This is a contradiction.
7 Theorems on the Index
If K≤H≤G and [G:K] is finite, then [G:K]=[G:H]⋅[H:K].
Let a1H,…,amH (m=[G:H]) be the left cosets of H in G, and let b1K,…,bnK (n=[H:K]) be the left cosets of K in H. Then G=i⨆aiH=i⨆aij⨆bjK=i,j⨆aibjK,
so {aibjK} is the complete set of left cosets of K in G, giving [G:K]=mn.To confirm that the cosets aibjK are all distinct: if aibjK=ai′bj′K, then aibj and ai′bj′ lie in the same K-coset. Thus aibjbj′−1ai′−1∈K⊆H, which gives aiH=ai′H, so i=i′. It then follows that bjK=bj′K, so j=j′. □
Let H,K≤G with [G:H] and [G:K] both finite. Then [G:H∩K]≤[G:H]⋅[G:K].
Equality holds when [G:H] and [G:K] are coprime.
The map ψ:G/(H∩K)→G/H×G/K defined by a(H∩K)↦(aH,aK) is well-defined and injective (if aH=bH and aK=bK, then a−1b∈H∩K). So [G:H∩K]≤[G:H][G:K].When gcd([G:H],[G:K])=1: from H∩K≤H and H∩K≤K, multiplicativity of the index gives [G:H]∣[G:H∩K] and [G:K]∣[G:H∩K]. Since these indices are coprime, [G:H][G:K]∣[G:H∩K]. Combined with the reverse inequality, equality follows. □
8 Right Cosets and Double Cosets
[G:H]L=[G:H]R (the number of left cosets equals the number of right cosets).
The map aH↦Ha−1 is a bijection from the set of left cosets to the set of right cosets. Indeed, aH=bH⟺a−1b∈H⟺b−1a∈H⟺Ha−1=Hb−1, so the map is well-defined and injective. Surjectivity is clear. □
A subgroup H≤G is called a normal subgroup, written H⊴G, if aH=Ha for all a∈G. When H is normal, left cosets and right cosets coincide, and the set of cosets can be endowed with a natural group structure (the quotient group).
9 Worked Examples
Consider the dihedral group D4={e,r,r2,r3,s,rs,r2s,r3s} of order 8, and its subgroup H=⟨r2⟩={e,r2} of order 2. The left cosets are: eHsH={e,r2},={s,r2s},rHrsH={r,r3},={rs,r3s}.
There are 4 cosets, so [D4:H]=4. Check: ∣D4∣=8=4⋅2=[D4:H]⋅∣H∣.
Since A4≤S4 with ∣A4∣=12 and ∣S4∣=24, we have [S4:A4]=2. The two cosets are A4 (the even permutations) and (12)A4 (the odd permutations).
[Z:nZ]=n. The cosets are 0+nZ,1+nZ,…,(n−1)+nZ, and they form the additive group Z/nZ.
10 Summary and Next Steps
In this chapter we have covered:
The definition of left and right cosets, and the fact that they partition the group.
Lagrange's theorem: ∣G∣=[G:H]⋅∣H∣.
Corollaries: the order of an element divides the order of the group; groups of prime order are cyclic.
Applications: Fermat's little theorem and Euler's theorem.
The converse of Lagrange's theorem fails in general (the A4 counterexample).
Multiplicativity of the index and the index of an intersection.
Lagrange's theorem tells us that the order of a subgroup is constrained, but it does not address whether a subgroup of every allowable order actually exists. Answering this converse question (at least partially) is the goal of the Sylow theorems, which require the theory of normal subgroups and quotient groups as a prerequisite.