Section 11: Normal Subgroups

Lehman MAT-314 Section 11: Normal Subgroups

1. Abstract Algebra, Saracino Second Edition Section 11 Normal Subgroups "In mathematics the art of proposing a question must be held of higher value than solving it." – Georg Cantor -

2. Galois and Abel Niels Henrik Abel (1802-1829) - Norwegian Mathematician Évariste Galois (1811-1832) - French Mathematician

3. Definition of Normal Subgroups (1 of 1 ) Definition: Let be a group. A subgroup of is said to be a normal subgroup of G if for all . Note: For any group , and are normal subgroups since for any , and . It follows directly from the definition that every subgroup of an abelian group is normal. However, the converse of the last statement is not true. We can show that every subgroup of the group of unit quaternions is normal, but itself is nonabelian. Note also that if is a normal subgroup of a group , then it does not always mean that for every and every . This will be demonstrated in the following example.  

4. Examples of Normal Subgroups (1 of 3 ) Example 1. Consider the group : Let . Then is a subgroup of . Since , , and are elements of , then , , and . Now, Hence, . Again, Therefore            

5. Examples of Normal Subgroups (2 of 3 ) Example 1 (cont’d). Similarly, we can show that It follows that is a normal subgroup of . But for and , we have: Hence, . Now, consider the subgroup of . For this subgroup: Hence, , and so is not a normal subgroup of          

6. Examples of Normal Subgroups (3 of 3 ) Example 2. Recall the group of real matrices with nonzero determinant. The set of matrices in with determinant is a subgroup of . First, is a nonempty subset of , because . Next, if then and . So is closed under both multiplication and inverses. In addition since if and then Above we used the fact that and from linear algebra.      

7. Normal Subgroups (1 of 8 ) Theorem 1 . Let be a group and let be a subgroup of . The following conditions are equivalent : is a normal subgroup of for all for all Proof . Let . Since is a normal subgroup of , then . Hence for any , , so for some . Consequently, we have , and . Let , then .  

8. Normal Subgroups (2 of 8 ) Theorem 1 (cont’d) . Let , then . This means that we have or . Hence, it follows that , and . Since for all , then for all . Hence, is a normal subgroup of . Theorem 11.2 . Let be a group, then any subgroup of the center is a normal subgroup of . Proof . Let be any subgroup of . Let and let . Since we have then and . It follows that and is normal in , by Theorem 1.  

9. Normal Subgroups (3 of 8 ) Example 2. Let be a group and let be a subgroup of . If for all , implies , show that is normal in . Solution . Let be arbitrary. Then for any Since implies for any , It follows that for all and for all . This means that and is normal in , by Theorem 1.    

10. Normal Subgroups (4 of 8 ) Corollary 1 . If is abelian then every subgroup of is normal. Proof . If is abelian then and every subgroup of is normal by Theorem 11.3. Theorem 11.3 . Let be a group and let be a subgroup of such that , then is normal in . ( index subgroups are normal ). Proof . Since there are exactly two distinct right cosets of in , one of them is and the other must be (set difference) since right cosets of partition . Likewise, the distinct left cosets must be and .  

11. Normal Subgroups (5 of 8 ) Proof (cont’d). Thus, the left cosets are the right cosets, and for any , we have . Example 3 . We know that every subgroup of an abelian group is normal. Is the converse true? That is, if every subgroup of a group is normal, is the group abelian? Solution . Consider the group of unit quaternions, where under multiplication, with . The subgroups of are: , , , , , and . It follows that is a normal subgroup of itself. The subgroups , , and all have order , so they are index subgroups and are therefore normal by Theorem 11.3.  

12. Normal Subgroups (6 of 8 ) Solution (cont’d). The subgroups and are both contained in the center and are therefore normal by Theorem 11.2. Thus, every subgroup of is normal, but itself is nonabelian. Example 4. We have and so the index and is a normal subgroup of for all by Theorem 11.3. Theorem 11.4 . Let be a group and let be a subgroup of . Suppose , then is a subgroup of with the same number of elements as . Proof. Let , then . Now,  

13. Normal Subgroups (7 of 8 ) Proof (cont’d). Now, since we have . We will now show that is closed under multiplication. Let and be elements of for some . Then since because is a subgroup of . For closure under inverses, let for some . Then since because is a subgroup of . It follows that is a subgroup of .        

14. Normal Subgroups (8 of 8 ) Proof (cont’d). To prove that we will construct a one-to-one and onto function from to . Let be given by for all . To show that is one-to-one, let and let then . By left and right cancellation, and is one-to-one. To show that is onto , let be an arbitrary element of , then for some . It follows that is onto , and Corollary 11.5 . If is a group and a subgroup of and no other subgroup has the same number of elements as , then is normal in .  

15. Quotient (Factor) Groups (1 of 5 ) Proof . For any , is a subgroup of with the same number of elements as . By the hypothesis, , so by Theorem 1, is normal in . Notation : If is a normal subgroup of a group , we write: Notation : If , then is defined as the set of all left (= right) cosets of in . That is Theorem 11.6 . Let be a group and let be a normal subgroup of , then is a group under the operation defined below: For all ,    

16. Quotient (Factor) Groups (2 of 5 ) Proof . We want to show that the operation is a well-defined binary operation on the set . That is, if we have and , then Now means and means . It follows that we want: That is, we want . But          

17. Quotient (Factor) Groups (3 of 5 ) Proof (cont’d) . We have Since and , then is an element of , if and only if is a subset of . That is, if is normal in . We will now check if is associative. Let be elements of , then Hence, is an associative binary operation on .          

18. Quotient (Factor) Groups (4 of 5 ) Proof (cont’d) . We will now check if has an identity element. Let be an element of , then So is the identity element in under . For inverses, let , then It follows that is closed under inverses and is a group under the binary operation . Definition : Let be a group and let be a normal subgroup of . Then the group of all left cosets of in under the binary operation is called the quotient group or factor group of by .      

19. Quotient (Factor) Groups (5 of 5 ) Definition : Let be a group and let be a normal subgroup of . Then the group of all left cosets of in under the binary operation is called the quotient group or factor group of by . We defined (the index of and ) as the number of distinct left (or right) cosets . Since is the set of all left (or right) cosets of a normal subgroup , then: If the group is finite, then by Lagrange’s Theorem, we have . It follows that for finite groups :      

20. Examples of Quotient Groups (1 of 3 ) Example 5 . Let under addition modulo . Then is cyclic, hence abelian. From Corollary 1, we know that every subgroup of is normal. Let be a subgroup of . Then , and the distinct left (right) cosets of in are: , , and . We see that , and with , so and are inverses of each other. and both have order This is actually the group        

21. Examples of Quotient Groups (2 of 3 ) Example 6 . In Example 1, we looked at We showed that the subgroup is normal in . Let us look at the set of distinct cosets of in : Here: We see that , and with , so Now, So is its own inverse. This is the group            

22. Examples of Quotient Groups (1 of 3 ) Example 7 . Consider the group under addition and consider the subgroup . Since is abelian, we know from Corollary 1 that every subgroup of is normal. The left cosets of in are: The distinct left cosets are: , and . Then                  

23. Corollaries (1 of 10 ) Example 8 . Let be a group and let be a normal subgroup of . Show that if is abelian then the quotient group is also abelian. Solution. Let and be arbitrary elements of for some . Then But for all since is abelian. Hence, Since for all in it follows that is abelian.      

24. Corollaries (2 of 10 ) Example 9 . Let be a group and let and be subgroups of with normal in . Show that is a normal subgroup of . Solution . We know that is a subgroup of . Since is contained in both and then is a subgroup of and , respectively. In order show that , it is sufficient to show that for all : Let , then and .Hence, for all . We want to show that . Since , then for all and all . But so . It follows that .    

25. Corollaries (3 of 10 ) Solution (cont’d) . Now, and for all . It follows that , and we have for all , but this implies that is normal in by Theorem 1. Example 10 . Let be an abelian group of odd order and let . Show that the product Solution. Since is of odd order, by Lagrange’s Theorem can have no element of order (the order of an element must divide the order of the group). Thus, each nonidentity element is distinct from its respective inverse. Since is abelian, then the above product is a product of copies of the identity.    

26. Corollaries (4 of 10 ) Theorem 3 . Let be a group. Show that if is cyclic, then is abelian. Proof : Let . We know that , so is a group under the operation for all elements . Since is cyclic, then for some . Let , then . Since is cyclic then and, also for some . It follows that we have and . Therefore, and for some . Now,      

27. Corollaries (5 of 10 ) Example 11 . Let be a group and let . Suppose . Show that Solution : Since , then for some , . Since , then for all , but So for all . Therefore for all . It follows that . Since , then Theorem 4 . Let be a finite group and let and be subgroups of .      

28. Corollaries (6 of 10 ) Theorem 4 . Let be a finite group and let and be subgroups of . If and are relatively prime ( , then . If and are two distinct subgroups, both of prime order , then . Proof (a). Now, is a subgroup of ( cf. Homework 5, #1 ). Since is contained in both and , then is subgroup of both and . By Lagrange’s Theorem, is a divisor of both and , but , so and .  

29. Corollaries (7 of 10 ) Proof (b). Now, is a subgroup of , which has prime order. By Lagrange’s Theorem, or . Since , then or . Since , if , then . But and , so there is some with . Therefore cannot be a subset of . It follows then that . Theorem 5. Let be a group of order for distinct primes and ( . Then all proper subgroups of are cyclic and has an element of order Proof . By Lagrange’s Theorem, the order of all subgroups of is a divisor of . The only proper  

30. Corollaries (8 of 10 ) Proof (cont’d). The only proper divisors of are , , and . So, if is a proper subgroup of , then we have , , or . If , then is cyclic. Otherwise, if , or (primes) then is cyclic. Let be the maximum order of all elements of . Then, by Lagrange’s Theorem , , , or . If , then has an element of order . Thus, is cyclic. Since , then and is an element of order in . If , then has an element of order . However, if then there exists of order . Consider the cyclic subgroup of generated by . . Every nonidentity element of has order .  

31. Corollaries (9 of 10 ) Proof (cont’d). From Theorem 4(b), if is another subgroup of of order , then . It follows that has elements of order that are not in , and vice versa . Since has one element of order , Then can have at most elements of order It follows that can have at most distinct subgroups of order . Thus, the maximum number of elements of order in is Since if had distinct subgroups of order , it would have elements of order . Now, . But , so and the number of elements of order would exceed .  

32. Corollaries (10 of 10 ) Proof (cont’d). But the total number of nonidentity elements in is . We know that has at most elements of order . So, the difference: Since the smallest prime is It follows that has at least one element of order .