Example. Let G be a finite group, H, K be subgroups of G. Then
|HK| = |H| · |K|/|H ∩ K|.
Quotient Groups
S
Proof. Since HK = {hk|h ∈ H, k ∈ K} = 1≤i≤m hi K, where
hi K ∩ hj K = ∅, ∀i 6= j, and for a, b ∈ H, aK = bK ⇔ a−1 bK =
K ⇔ a−1 b ∈ K ⇔ a−1 b ∈ K ∩ H ⇔ a(K ∩ H) = b(K ∩ H). This
implies the number of cosets of K in HK is equal to the number
of cosets of K ∩ H in H. So |HK| = |H| · |K|/|H ∩ K|.
Quotient Groups
Quotient Groups
Xiamen University
03/30/2015
Quotient Groups
Quotient groups
Quotient Groups
Quotient groups
Lemma. Let N C G, then the product of the cosets aN, bN
for a, b ∈ G is again a cosets. In fact, aN · bN = (ab)N .
Quotient Groups
Quotient groups
Lemma. Let N C G, then the product of the cosets aN, bN
for a, b ∈ G is again a cosets. In fact, aN · bN = (ab)N .
Notation. We set G/N = {aN |a ∈ G}.
Quotient Groups
Quotient Groups
Theorem. Let N C G, then with the multiplication defined as
above, the set G/N forms a group, and the map
π : G → G/N
a 7→ aN
is surjective group homomorphism with kernel kerπ = N, and
the order of the group G/N is equal to the index [G : N ].
Quotient Groups
Theorem. Let N C G, then with the multiplication defined as
above, the set G/N forms a group, and the map
π : G → G/N
a 7→ aN
is surjective group homomorphism with kernel kerπ = N, and
the order of the group G/N is equal to the index [G : N ].
Corollary. Every normal subgroup N of a group G is the
kernel of some group homomorphism.
Quotient Groups
Quotient Groups
Lemma. Let G be a group. S be a set with a law of
composition. Let ϕ : G → S be a surjective map such that
ϕ(ab) = ϕ(a)ϕ(b), ∀a, b ∈ G. Then S is a group.
Quotient Groups
Lemma. Let G be a group. S be a set with a law of
composition. Let ϕ : G → S be a surjective map such that
ϕ(ab) = ϕ(a)ϕ(b), ∀a, b ∈ G. Then S is a group.
Theorem (First Isomorphism Theorem). Let ϕ : G → G1 be
a surjective group homomorphism, and N =kerϕ. Then
G/N ∼
= G1 , and the group isomorphism is given by
ϕ̄ : G/N → G1
aN 7→ ϕ(a).
Quotient Groups
Quotient Groups
Corollary. (1) If ϕ : G → G1 is a group homomorphism, then
G/kerϕ ∼
= Imϕ.
Quotient Groups
Corollary. (1) If ϕ : G → G1 is a group homomorphism, then
G/kerϕ ∼
= Imϕ.
(2) If ϕ : G → G1 is an injective homomorphism, then
G∼
= Imϕ.
Quotient Groups
Quotient Groups
Example. If Z is the center of the group G, and suppose that
G/Z is a cyclic group. Then G is an abelian group.
Quotient Groups
Example. If Z is the center of the group G, and suppose that
G/Z is a cyclic group. Then G is an abelian group.
Example. If ϕ : G → H is a surjective homomorphism of
groups, and N C H, prove
(1) ϕ−1 (N ) C G.
(2) G/ϕ−1 (N ) ∼
= H/N .
Quotient Groups
Second and Third Isomorphism Theorems
Quotient Groups
Second and Third Isomorphism Theorems
Let H, K be subgroups of G, and K C G. Then
HK/K ∼
= H/H ∩ K.
Quotient Groups
Second and Third Isomorphism Theorems
Let H, K be subgroups of G, and K C G. Then
HK/K ∼
= H/H ∩ K.
Let N, M be normal subgroups of G, and N ⊂ M . Then
(G/N )/(M/N ) ∼
= G/M . (Hw.)
Quotient Groups
Quotient Groups
Example. Let G be a finite group, N C G, and |N | and
|G/N | be coprime. If a ∈ G and the order of a divides |N |,
then a ∈ N .
Quotient Groups
Homework
Quotient Groups
Homework
1.P74.10.2. Let H and K be subgroups of a group G.
(a) Prove that the intersection xH ∩ yK of two cosets of H
and K is either empty or a coset of the subgroup H ∩ K.
(b) Prove that if H and K have finite index in G then H ∩ K
also has finite index in G.
2.P74.11.6. Let G be a group that contains normal
subgroups of orders 3 and 5, respectively. Prove that G
contains an element of order 15.
3.P74.12.1. Show that if a subgroup H of a group G is not
normal there are left cosets aH and bH whose product is not
a coset.
4. Let N, M be normal subgroups of G, and N ⊂ M . Prove
that (G/N )/(M/N ) ∼
= G/M .
Quotient Groups