Chapter 4
Cosets and Lagrange’s
theorem
We begin by revisiting what we saw when building Z/nZ.
We introduced an equivalence relation between elements of Z:
x ∼ y ⇔ x − y ∈ nZ.
(4.1)
And we saw that the equivalence class of x ∈ Z for this equivalence relation is:
x = {. . . , x − 2n, x − n, x, x + n, x + 2n, x + 3n, . . .}
= x + nZ.
Observe also that nZ is a subgroup of Z (all the subgroups of Z are of this form,
see some exercise sheet).
We want to extend this to arbitrary groups:
Definition 4.1. Let G be a group and let H be a subgroup of G. We define an
equivalence relation ∼H between elements of G by
x ∼H y ⇔ xy −1 ∈ H.
We write x = y mod H if x ∼H y and say that x is equal to y modulo H.
Observe that the definition of ∼H follows exactly the pattern of (4.1), we
are just using H instead of nZ and a multiplicative notation for the group
operation:
For a group (G, +) where the operation is denoted by the symbol + (so for
instance (Z, +)) we would write
x ∼H y ⇔ x − y ∈ H.
Lemma 4.2. The relation ∼H is an equivalence relation.
Proof. We check the three properties of equivalence relations:
We have x ∼H x because xx−1 = e ∈ H.
Assume x ∼H y, i.e. xy −1 ∈ H. Since H is a subgroup, we have (xy −1 )−1 ∈ H,
i.e. yx−1 ∈ H, in other words y ∼H x.
Finally assume x ∼H y and y ∼H z, i.e. xy −1 ∈ H and yz −1 ∈ H. Since H is a
subgroup we have (xy −1 )(yz −1 ) ∈ H, i.e. az −1 ∈ H, which gives x ∼H z.
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CHAPTER 4. COSETS AND LAGRANGE’S THEOREM
Proposition 4.3. The equivalence class of a ∈ G for the relation ∼H is:
a = Ha,
where aH is defined as follows
Ha := {xa | x ∈ H}.
Proof. We want to show an equality between two sets. One way to do this is to
show that they contain exactly the same elements, i.e. that each set is contained
in the other.
We start with b ∈ a and show that b ∈ aH. By definition of a we have
ab−1 ∈ H, so there is x ∈ H such that ab−1 = x. We solve for b: Multiplying
both sides on the right by b give a = xb, and on the left by x−1 gives b = x−1 a,
which belongs to aH since x−1 ∈ H.
Take now b ∈ Ha, i.e. b = xa for some x ∈ H. Then ab−1 = aa−1 x−1 =
−1
x ∈ H, so a ∼H b, i.e. b ∈ a.
Definition 4.4. Let H be a subgroup of G. A set of the form Ha, for somce
a ∈ G, is called a right coset of H. A left coset of H is a set of the form
Ha.
Remark 4.5.
1. In general we do not have Ha = aH. See exercise sheets
for an easy counter-example.
2. We could define another equivalence relation on G by x ∼0H y if and
only if a−1 b ∈ H. The equivalence class of an element a would then be
aH = {ax | x ∈ H}. This whole chapter can actually be written using this
equivalence relation and left cosets instead of right cosets.
The next result is the key to Lagrange’s theorem, and we have already proved
most of it.
Proposition 4.6.
1. Assume H is finite and let a ∈ G. Then |Ha| = |H|.
2. Every element of G belongs to a right coset of H.
3. Any two right cosets of H are equal or have no element in common.
Proof.
1. To show that H and Ha have the same number of elements we
define a bijection between them:
f : H → Ha, x 7→ xa.
f is injective: Assume that xa = ya. Then, multiplying on the right by
a−1 we obtain x = y.
f is surjective: Every element of Ha is in the image of f by definition of
f.
2. We know that x ∼H x since ∼H is an equivalence relation, which means
x ∈ x = Hx.
3. This is a consequence of Proposition 2.8 since the right cosets of H are
equivalence classes.
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Theorem 4.7 (Lagrange’s Theorem). Let G be a finite group and let H be a
subgroup of G. Then |H| divides G|.
Proof. By Proposition 4.6, the group G is “paved” by the right cosets of H:
Every element belongs to one, and they do not overlap. In particular there are
only finitely many different right cosets of H: Ha1 , . . . , Hak , and a picture
would look like this:
Ha1
Ha2
Ha3
Ha4
G
Hak
Counting the elements of G, we obtain
|G| = |Ha1 | + |Ha2 | + · · · + |Hak |
= |H| + |H| + · · · + |H|
by Proposition 4.6
= k|H|.
Remark 4.8.
1. One of the remarkable aspects of this result is that the definitions of group and subgroup are algebraic conditions that do not appear
to require anything about the number of elements. But they actually do.
2. Lagrange’s theorem was not proved by Lagrange. The notion of group did
not even exist at the time. But he was the first (or one of the first) to
come up with similar ideas in the domain that would later give rise to the
notion of group (the study of solutions of polynomial equations).
Definition 4.9. Let G be a group and let H be s subgroup of G. The index of
H in G, denoted [G : H], is the number of distinct right cosets of H in G.
Remark 4.10. So by Lagrange’s theorem (or more precisely its proof ):
[G : H] = |G|/|H|.
The set of all right cosets of H in G is often denoted by G/H:
G/H = {Ha | a ∈ G},
so [G : H] = |G/H| = |G|/|H|.
Corollary 4.11. Let G be a finite group and let a ∈ G. Then o(a) divides |G|.
Proof. We saw in Proposition 3.27 that o(a) = |hai|, and the result follows by
Lagrange’s theorem since hai is a subgroup of G.
Corollary 4.12. Let G be a finite group and let a ∈ G. Then a|G| = e.
Proof. By the Corollary 4.11 |G| = k · o(a) for some k ∈ N. Therefore
a|G| = ako(a) = (ao(a) )k = ek = e.
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CHAPTER 4. COSETS AND LAGRANGE’S THEOREM
Theorem 4.13. Let G be a group such that |G| is a prime number. Then G is
a cyclic group.
Proof. Let p = |G|. Let a ∈ G such that a 6= e (it exists since |G| = p ≥
2). Consider hai, the subgroup generated by a. It contains at least 2 dictinct
elements: e and a, so |hai| ≥ 2. But |hai| divides p, so, since p is prime, we must
have |hai| = p = |G|, which gives hai = G.
Observe that what we proved is slightly stronger than the conclusion of the
theorem: G is generated by any one of its elements that is not e.