# Abstract Algebra/Group Theory/Normal subgroups and Quotient groups

In the preliminary chapter we discussed equivalence classes on sets. If the reader has not yet mastered this notion, he/she is advised to do so before starting this section.

## Normal Subgroups[edit]

Recall the definition of kernel in the previous section. We will exhibit an interesting feature it possesses. Namely, let be in the coset . Then there exists a such that for all . This is easy to see because a coset of the kernel includes all elements in that are mapped to a paticular element. The kernel inspires us to look for what are called *normal subgroups*.

**Definition 1:** A subgroup is called *normal* if for all . We may sometimes write to emphasize that is normal in .

**Theorem 2:** A subgroup is normal if and only if for all .

*Proof*: By the definition, a subgroup is normal if and only if since conjugation is a bijection. The theorem follows by multiplying on the right by . ∎

We stated that the kernel is a normal subgroup in the introduction, so we had better well prove it!

**Theorem 3:** Let be any homomorphism. Then is normal.

*Proof*: Let and . Then , so , proving the theorem. ∎

**Theorem 4:** Let be groups and a group homomorphism. Then if is a normal subgroup of , then is normal in .

*Proof*: Let and . Then since is normal in , and so , proving the theorem. ∎

**Theorem 5:** Let be groups and a group homomorphism. Then if is a normal subgroup of , is normal in .

*Proof*: Let And . Then if such that , we have for some since is normal. Thus for all and so is normal in . ∎

**Corollary 6:** Let be groups and a surjective group homomorphism. Then if is a normal subgroup of , is normal in .

*Proof*: Replace with in the proof of Theorem 5. ∎

**Remark 7:** If is a normal subgroup of and is a normal subgroup of , it does **not** necessarily imply that is a normal subgroup of . The reader is invited to display a counterexample of this.

**Theorem 8:** Let be a group and be subgroups. Then

- i) If is normal, then is a subgroup of .

- ii) If both and are normal, then is a normal subgroup of .

- iii) If and are normal, then is a normal subgroup of .

*Proof*: i) Let be normal. First, since for each , there exists such that , so . To show is a subgroup, let . Then for some since is normal, and so is a subgroup.

ii) Let and . Then since both and are normal, there exists such that . It follows that and so is normal.

iii) Let and . Then since H is normal, and similarly . Thus and it follows that is normal. ∎

### Examples of Normal Subgroups[edit]

In the following, let be any group. Then has associated with it the following normal subgroups.

- i) The
*center*of , denoted , is the subgroup of elements which commute with all others. . That is a normal subgroup is easy to verify and is left to the reader.

- ii) The
*commutator subgroup*of , denoted or , is the subgroup generated by the subset where for all . For , we introduce the shorthand . Then we have , such that for any product of commutators where all elements are in , we have , and so is normal.

**Remark 9:** We can iterate the commutator subgroup construction and define and for all . We will not use the commutator subgroup in future results in this book, so for us it is merely a curiosity.

## Equivalence Relations on Groups[edit]

Why are normal subgroups important? In the preliminary chapter we discussed equialence relations and the associated set of equivalence classes. If is a group and is an equivalence relation, when does admit a group structure? Of course we have to specify the multiplication on . We will do so now.

**Definition 10:** Let be a group and is an equivalence relation on , we define multiplication on the equivalence classes in such that for all ,

This is indeed the only natural way to do it. Take the two equivalence classes, choose representatives, compute their product and take its equivalence class. The alert reader will have only one thing on his/her mind: is this well defined? For a general equivalence relation, the answer is no. The reader is invited to come up with an example. What is more interesting is *when* is it well defined? By the definition above, we obviously need the projection map defined by to be a homomorphism. We can in fact condense the requirements down to two, both having to do with cancellation laws.

**Theorem 11:** Let be a group and an equivalence relation on . Then is a group under the natural multiplication if and only if for all

- .

*Proof*: Assume is a group. Since , the property follows from the cancellation laws in . Assume now that the property holds. Then its multiplication rule is well defined, and must verify that is a group. Let , then associativity is inherited from :

- .

The identity in is the equivalence class of , :

- .

Finally, the inverse of is :

- .

So really defines a group structure, proving the theorem. ∎

We will call an equivalence relation *compatible* with if is a group. Then, is called the *quotient group* of by . Also, as an immediate consequence, this makes into a homomorphism, but not just any homomorphism! It satisfies a universal property!

**Theorem 12:** Let be en equivalence relation compatible with , and a group homomorphism such that . Then there exists a unique homomorphism such that .

*Proof*: In the preliminary chapter on set theory, we showed the corresponding statement for sets, so we know that exists as a function between sets. We have to show that it is a homomorphism. This follows immediately: since by commutativity, we have . As stated already, shows uniqueness, proving the theorem. ∎

**Lemma 13:** Let be an equivalence relation on a group such that . Then is a subgroup of and .

*Proof*: First off, is nonempty since . Let . Then by multiplying on the left by . Then since we have by the same argument. Applying transitivity gives . Finally, multiplying on the left by gives , giving and so is a subgroup.

Assume for . Then implying . Thus . Now assume . Then and so and finally .

Assume . Then since is a subgroup, we have and so . Finally, assume . Then . Since in paticular , this implies , completing the proof. ∎

The mirror version using right cosets and the equivalence relation and is completely analogous. Stating the theorem and writing out the proof is left to the reader as an exercise.

We have showed how an equialence relation defines a subgroup of . In fact the equivalence classes are all cosets of this subgroup. We will now go the other way, and show how a subgroup defines an equivalence relation on .

**Lemma 14:** Let be a subgroup of a group . Then,

- i) is an equivalence relation such that for all .

- ii) is an equivalence relation such that for all .

*Proof*: We will prove i). The proof for ii) is similar and is left as an exercise for the reader.

The fact that is an equivalence relation and that was proven in the section on subgroups. Assume . Then for all , such that . Now assume , Then such that , completing the proof. ∎

**Theorem 15:** For every equivalence relation on G such that , there exists a unique subgroup of such that are precisely the left cosets of .

*Proof*: This follows from Lemma 13 and Lemma 14.

Again, the mirror statement is completely analogous. Stating the theorem is left to the reader as an exercise.

## Quotients with respect to Normal Subgroups[edit]

**Lemma 16:** Let be the equivalence relation given by , where is a subgroup of G. Then we know that is compatible if and only if is a normal subgroup.

*Proof*: Assume is compatible, and . Then , and compatibility gives us , and so . Since is arbitrary, we obtain for all and so is normal. Assume now that is normal. Then , and for all . Using this, we obtain and similarily for the right hand case, so is compatible with . ∎

**Definition 17:** When an equivalence relation is given by specifying a normal subgroup , the quotient group with respect to this equivalence relation is denoted . We then refer to as the quotient of with respect to , or modulo . Note that this complies with previous definitions of this notation.

Multiplication in is given as before as , with identity and for all .

**Definition 18:** Let be a normal subgroup of . Then we define the *projection homomorphism* by for all .

**Theorem 19:** A subgroup is normal if and only if it is the kernel of some homomorphism.

*Proof*: We have already covered the left implication. For the right implication, assume is normal. Then is a group and we have the projection homomorphism as defined above. Since for all we have , and so is the kernel of a homomorphism. ∎

**Theorem 20:** Let be groups, a homomorphism and a normal subgroup of such that . Then there exists a unique homomorphism such that .

*Proof*: This follows from Theorem 12 by letting . ∎

## The Isomorphism Theorems[edit]

**Theorem 21 (First Isomorphism Theorem):** Let be groups and a homomorphism. Then .

*Proof*: From Theorem 20 we have that there exists a unique homomorphism such that . We have to show that is an isomorphism when we corestrict to . This is immediate, since by Lemma 13, so that is injective, and for any there is a such that so that it is surjective and therefore an isomorphism. ∎

**Lemma 22:** Let be a group, a subgroup and a normal subgroup of . Then is a normal subgroup of .

*Proof*: Let and . Then since and is a subgroup and since , and is normal in . Thus and is normal in . ∎

**Theorem 23 (Second Isomorphism Theorem):** Let be a group, a subgroup and a normal subgroup of . Then .

*Proof*: Define by for all . is surjective since any element in can be written as with and , so . We also have that and so by the first isomorphism theorem. ∎

**Lemma 24:** Let be a group, and let be normal subgroups of such that . Then is a normal subgroup of .

*Proof*: Let and . Then for some since is normal. Thus , showing that is normal in . ∎

**Theorem 25 (Third Isomorphism Theorem)** Let be a group, and let be normal subgroups of such that . Then .

*Proof*: Let be given by . This is well defined and surjective since , and is a homomorphism. Its kernel is given by , so by the first isomorphism theorem, . ∎

**Theorem 26 (The Correspondence Theorem):** Let be a group and be a normal subgroup. Now let and . Then is an order-preserving bijection from to .

*Proof*: We must show injectivity and surjectivity. For injectivity, note that if , then , so if such that , then , proving injectivity. For surjectivity, let . Then , so that , and , proving surjectivity. Lastly, since implies that , the bijection is order-preserving. ∎

**Note 27:** The correspondence Theorem is sometimes called *The Forth Isomorphism Theorem*.

**Theorem 28:** Let from Theorem 26. Then is normal if and only if is normal in , and then .

*Proof*: Since is surjective, normal implies normal. Assume that is normal. Then and so is normal since it is the preimage of a normal subgroup. To show the isomorphism, let be given by a composition of projections: . Then , so by the first isomorphism teorem, . ∎

**Corollary 29:** Let be a group and a normal subgroup. Then for any there exists a unique subgroup such that and . Also, is normal in if and only if is normal in .

*Proof*: From Theorem 26 we have that the projection is a bijection, and since for all , we have . The second part follows from Theorem 28. ∎

## Simple Groups[edit]

**Definition 30:** A group is called *simple* is it has no non-trivial proper normal subgroups.

**Example 31:** Every cyclic group , where is prime, is simple.

**Definition 32:** Let be a group and a normal subgroup. is called a *maximal normal subgroup* if for any normal subgroup of , we have .

**Theorem 33:** Let be a group and a normal subgroup. Then is a maximal normal subgroup if and only if the quotient is simple.

*Proof*: By Theorem 26 and Theorem 28, has a nontrivial normal subgroup if and only if there exists a proper normal subgroup of such that . That is, is not maximal if ans only if is not simple. The theorem follows. ∎

## Problems[edit]

**Problem 1:** Recall the unitary and special unitary groups from the section about subgroups. Define the *projective unitary group of order* as the group . Similarily, define the *projective special unitary group of order* as .

- i) Show that

- ii) Using the second isomorphism theorem, show that .