# Quotient groups

One of the most useful facts about normal subgroups is the following result/definition.

Lemma 5.18.1   If is a normal subgroup of then the coset space with the binary operation,

for all belonging to , is a group. The identity element of this group is the trivial coset .

This group is called the quotient group of by and is sometimes pronounced  mod ''.

Example 5.18.2   If is a homomorphism between two groups then is a group.

Next we introduce an important idea first emphasized by E. Galois (October 1811-May 1832), the French mathematician mentioned in the introduction. Though Galois attended secondary school, he had trouble entering the French university system. Sadly, by the time he entered the École Normale Súperieure in November 1829, he also began to get caught up in the activities of the French revolution which toppled Charles X. His political activities resulted in him being expelled from schoolin December 1830. Though apparently he had been a rather rebellious teenager, this only frustrated him further. He died in a fight with another Frenchman.

Galois's work was the first to begin to illuminate the basic building blocks'' of the collection of finite groups (i.e., the analog of the idea that atoms are the basic building blocks'' of molecules). Galois introduced the ideas of solvable groups and normal groups. Roughly speaking, one may say that the basic building blocks of finite groups are those the groups which have no proper non-trivial normal subgroups. Intuitively, this is because a non-trivial quotient group (by a normal subgroup) is closely related to the original group but smaller in size (and hence perhaps subject to analysis by an inductive argument of some type). These basic building blocks are called simple'' groups.

Definition 5.18.3   A simple group is a group with no proper normal subgroups other than the trivial subgroup .

All finite simple groups have been classified. In fact, the survey book discussing this is free and download able from [GLS].

Example 5.18.4   If is a prime then (the cyclic group having elements) is simple. In fact, if is any group which is both abelian and simple then there is a prime such that . If then is simple (as was stated above in Theorem 5.17.7). These facts are proven in [R].

Simple groups are not very abundant. In fact, the first non-abelian simple group is of order (it's ).

Example 5.18.5   Let denote the map which sends an integer to its residue . This map send every multiple of to . In fact, if then . Therefore, .

The following basic result, which generalizes the above example, describes the quotient group .

Theorem 5.18.6 (First Isomorphism Theorem)   If is a homomorphism between two groups then is isomorphic to .

proof: is a normal subgroup of , so is a group. We must show that this group is isomorphic to the group . Define by , for . We must show

(a) is well-defined,

(b) is a homomorphism,

(c) is a bijection.

If then , since is a group. This implies , so . This implies is well-defined.

Since is normal, . Therefore , for all . This implies is a homomorphism.

It is clear that is surjective. To show that is a bijection, it suffices to prove is an injection. Suppose that , for some . Then , so . By definition of the kernel, this implies , so . This implies is injective.

Example 5.18.7   If is the mod map'' then is an isomorphism.

In particular, if and are relatively prime integers then .

Indeed, by replacing by if necessary, we may assume that and are relatively prime integers. It suffices to show that the kernel is trivial since both and have the same cardinality. If sends to (where ) then , so since are relatively prime.

proof:

Example 5.18.8   Let denote the map which sends an integer to its residue . Let denote the map which sends an integer to its residue , where , are integers. Let , , so that .

Example 5.18.9   Let denote the map which sends an integer to its residue . Let denote the map which sends an integer to its residue , where , are integers. Let , , so that .

Then is a subgroup of and the map defined by sending to has kernel . By the first isomorphism theorem, we therefore have

Exercise 5.18.10   Let act on the set by permutations in the usual way. Let be arbitary.

(a) Describe the elements in the stabilizer as permutations in array form (recall (5.3) was the definition). Is this a normal subgroup of ?

(b) Compute the orbit .

(c) Show that there is a 1-1 correspondence between the elements in the orbit and elements in . (Hint: See the proof of Proposition 5.13.2.)

Exercise 5.18.11   Let act on the set by permutations in the usual way. Let denote the set of all -tuples of . The group acts on as well: if and then we define . Show that the stabilizer is a normal subgroup of if and only if has the form for some . In this case, compute the quotient group .

Exercise 5.18.12   Find all the normal subgroups of . For each one, compute .

david joyner 2008-04-20