Quotient Group: Definition, Properties, Solved Examples - Mathstoon (2024)

In this article, we will learn about the definition of quotient groups along with their examples, properties, and a few solved problems.

Definition of Quotient Group

Let H be a normal subgroup of a group G. Consider the set of all left cosets of H in G, that is, {aH: a∈G}. This set forms a group under the composition aH*bH = abH for all a,b ∈ G, denoted by G/H. This group (G/H, *) is called the quotient group of G by H.

Examples of Quotient Groups

Below are a few examples of quotient groups (or factor groups).

1. Let G=(Z₆, +) be the additive group and H={[0], [3]}. Then H is a normal subgroup of G and the quotient group G/H contains three elements H, [0]+H, [1]+H.

2. Let G=(Z, +) be the additive group of integers, and H=(2Z, +) is a normal subgroup of G. Then the set G/H consisting of left cosets of H in G forms the quotient group having two elements H and 1+H.

Properties of Quotient Groups

A quotient group satisfies the following properties.

1. The trivial group {e_G} is always a normal subgroup of G and the corresponding quotient group G/{e_G} is the group G itself.
2. The quotient group of G by G, that is, G/G is the trivial group.
3. The identity element in a quotient group G/H is the normal subgroup H itself.
4. The order of a quotient group G/H is equal to the order of G divided by the order of H. In other words, |G/H| = |G|/|H|.
5. If H is a subgroup of an abelian group G, then the quotient group G/H must be also abelian. But, not that the converse is not always true. For example, let G=S₃ be the permutation group on three objects and H=A₃ be its alternating group. As G/H =2, the quotient group G/H is abelian, but we know that G is not abelian.
6. If H is a subgroup of a cyclic group G, then the quotient group G/H is cyclic. But, not that the converse is not always true. In the above example, G/H=S₃/A₃ is cyclic but G=S₃ is not cyclic.
7. If G is a non-abelian group with center Z(G), then the factor group G/Z(G) is non-cyclic.
8. If G is a solvable group, then every quotient group G/H is also solvable.
9. The quotient group G/H is nilpotent if G is so.

Solved Problems on Quotient Group

Question 1: Prove that the symmetric group S₃ has a trivial center.

Answer:

Let Z denote the center of S₃. As S₃ is non-abelian, we conclude that Z≠S₃.

We know that Z is a normal subgroup of S₃. Thus, by Lagrange’s theorem, the order |Z| must be a divisor of |S₃|=6. As |Z|≠6, we deduce that

|Z| = 1, 2, or 3.

If |Z|=3, then |S₃/Z|=2. Hence, S₃/Z is cyclic but this is an impossibility by the above property 7. Therefore, |Z| cannot be equal to 3.

If |Z|=2, then |S₃/Z|=3. Hence, S₃/Z is cyclic. But, again this is an impossibility by the above property 7. Therefore, |Z| cannot be equal to 2.

Therefore, we obtain the conclusion that |Z|=1. Hence, S₃ has the trivial center.

Related Topics:

Group Theory: Definition, Examples, Orders, Types, Properties, Applications

Abelian Group: Definition, Properties, Examples

Cyclic Group: Definition, Orders, Properties, Examples

Normal Subgroup

Center of a Group

FAQs on Quotient Group