2.3. COSETS AND THE THEOREM OF LAGRANGE
2.3
29
Cosets and the Theorem of Lagrange
We always assume that H is a subgroup of the group G.
2.3.1
Cosets
Def 2.33. Let H be a subgroup of G. Given a ∈ G, the subset aH =
{ah | h ∈ H} of G is the left coset of H containing a, while the subset
Ha = {ha | h ∈ H} is the right coset of H containing a.
Ex 2.34. H = eH = He is both a left coset and a right coset (In general
aH 6= Ha).
Properties of left cosets: (Similar for right cosets)
1. The number of elements in aH is equal to the number of elements in
H.
2. If aH ∩ bH 6= ∅, then aH = bH.
(Proof: If aH ∩ bH 6= ∅, then ah1 = bh2 for some h1 , h2 ∈ H. So
−1
b = ah1 h−1
2 and bH = a(h1 h2 H) = aH.)
3. From 2., G can be partitioned into left cosets of H.
Prop 2.35 (Abelian Group). Let hG, +i be an abelian group and H ≤ G.
Then
1. a + H = {a + h | h ∈ H} = {h + a | h ∈ H} = H + a. So
“the left coset of H containing a” = “the right coset of H containing a”.
2. (a + H) + (b + H) = (a + b) + H. So the cosets of H have a “sum”
operation, which makes the set of all cosets of H a group (coset group).
Caution: If G is not abelian, the above conclusions may be wrong!
Ex 2.36 (Ex 10.3, p.97). The left cosets and right cosets of the subgroup
3Z ≤ Z.
Ex 2.37. When n ≥ 2, the coset of An in Sn is An and τ An for τ = (1, 2).
30
CHAPTER 2. PERMUTATIONS, COSETS, DIRECT PRODUCTS
2.3.2
The Theorem of Lagrange
Thm 2.38 (Theorem of Lagrange). Let H be a subgroup of a finite group
G. Then the order of H divides the order of G.
Proof. By the properties of cosets, the group G can be partitioned into left
cosets a1 H, a2 H, · · · , ak H that have empty intersection pairwise. So
|G| = |a1 H| + |a2 H| + · · · + |ak H| = |H| + |H| + · + |H| = k|H|.
|
{z
}
k copies
Therefore, |H| is a divisor of |G|.
Cor 2.39. The order of an element of a finite group divides the order of the
group.
Cor 2.40. Every group of prime order p is cyclic.
Therefore, if |G| = p where p is a prime, then G ' Zp .
Def 2.41. The number of left cosets of a subgroup H in a group G is the
index (G : H) of H in G.
Thm 2.42. Suppose H and K are subgroups of a group G such that K ≤
H ≤ G, and suppose that (H : K) and (G : H) are both finite. Then
(G : K) = (G : H)(H : K) is finite.
2.3.3
Homework, II-10, p.101-p.104
4, 12, 19, 38, 40.
(opt) 16, 28, 37, 45,