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 particular 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 equivalence 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!

 
Commutative diagram showing the universal property satisfied by the projection homomorphism.

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 particular  , 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 equivalence 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 similarly 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

 
Commutative diagram showing the first isomorphism theorem.   is an isomorphism.

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.

Theorem 2.6.? (Baumslag):

Let   be a group, let   be subgroups of   such that  , and let   be a subgroup of   such that  . Then:

 

Proof:

Due to theorem 2.6.?,   and   are subgroups of  . Further, theorem 2.6.? implies that  . Therefore, the function

 

is a homomorphism.

Further, since   is a subgroup of  , for all   we have:

 

And thus:

 

Therefore,  . Thus, the first isomorphism theorem implies

   

Simple Groups edit

Definition 30: A group is called simple if 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 and 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  . Similarly, define the projective special unitary group of order   as  .

i) Show that  
ii) Using the second isomorphism theorem, show that  .