Abstract Algebra/Group Theory/Subgroup

SubgroupsEdit

We are about to witness a universal aspect of mathematics. That is, whenever we have any sort of structure, we ask ourselves: does it admit substructures? In the case of groups, the answer is yes, as we will immediately see.

Definition 1: Let   be a group. Then, if   is a subset of   which is a group in its own right under the same operation as  , we call   a subgroup of   and write  .

Example 2: Any group   has at least 2 subgroups;   itself and the trivial group  . These are called the improper and trivial subgroups of  , respectively.

Naturally, we would like to have a method of determining whether a given subset of a group is a subgroup. The following two theorems provide this. Since   naturally inherits the associativity property from  , we only need to check closure.

Theorem 3: A nonempty subset   of a group   is a subgroup if and only if

(i)   is closed under the operation on  . That is, if  , then  ,
(ii)  ,
(iii)   is closed under the taking of inverses. That is, if  , then  .

Proof: The left implication follows directly from the group axioms and the definition of subgroup. For the right implication, we have to verify each group axiom for  . Firstly, since   is closed, it is a binary structure, as required, and as mentioned,   inherits associativity from G. In addition,   has the identity element and inverses, so   is a group, and we are done.

There is, however, a more effective method. Each of the three criteria listed above can be condensed into a single one.

Theorem 4: Let   be a group. Then a nonempty subset   is a subgroup if and only if  .

Proof: Again, the left implication is immediate. For the right implication, we have to verify the (i)-(iii) in the previous theorem. First, assume  . Then, letting  , we obtain  , taking care of (ii). Now, since   we have   so   is closed under taking of inverses, satisfying (iii). Lastly, assume  . Then, since  , we obtain  , so   is closed under the operation of  , satisfying (i), and we are done.

All right, so now we know how to recognize a subgroup when we are presented with one. Let's take a look at how to find subgroups of a given group. The next theorem essentially solves this problem.

Theorem 5: Let   be a group and  . Then the subset   is a subgroup of  , denoted   and called the subgroup generated by  . In addition, this is the smallest subgroup containing   in the sense that if   is a subgroup and  , then  .

Proof: First we prove that   is a subgroup. To see this, note that if  , then there exists integers   such that  . Then, we observe that   since  , so   is a subgroup of  , as claimed. To show that it is the smallest subgroup containing  , observe that if   is a subgroup containing  , then by closure under products and inverses,   for all  . In other words,  . Then automatically   since   is a subgroup of  .

Theorem 6: Let   and   be subgroups of a group  . Then   is also a subgroup of  .

Proof: Since both   and   contain the identity element, their intersection is nonempty. Let  . Then   and  . Since both   and   are subgroups, we have   and  . But then  . Thus   is a subgroup of  .

Theorem 6 can easily be generalized to apply for any arbitrary intesection   where   is a subgroup for every   in an arbitrary index set  . The reasoning is identical, and the proof of this generalization is left to the reader to formalize.

Definition 7: Let   be a group and   be a subgroup of  . Then   is called a left coset of  . The set of all left cosets of   in   is denoted  . Likewise,   is called a right coset, and the set of all right cosets of   in   is denoted  .

Lemma 8: Let   be a group and   be a subgroup of  . Then every left coset has the same number of elements.

Proof: Let   and define the function   by  . We show that   is a bijection. Firstly,   by left cancellation, so   is injective. Secondly, let  . Then   for some   and  , so   is surjective and a bijection. It follows that  , as was to be shown.

Lemma 9: The relation   defined by   is an equivalence relation.

Proof: Reflexivity and symmetry are immediate. For transitivity, let   and  . Then  , so   and we are done.

Lemma 10: Let   be a group and   be a subgroup of  . Then the left cosets of   partition  .

Proof: Note that   for some  . Since   is an equivalence relation and the equivalence classes are the left cosets of  , these automatically partition  .

Theorem 11 (Lagrange's theorem): Let   be a finite group and   be a subgroup of  . Then  .

Proof: By the previous lemmas, each left coset has the same number of elements   and every   is included in a unique left coset  . In other words,   is partitioned by   left cosets, each contributing an equal number of elements  . The theorem follows.

Note 12: Each of the previous theorems have analagous versions for right cosets, the proofs of which use identical reasoning. Stating these theorems and writing out their proofs are left as an exercise to the reader.

Corollary 13: Let   be a group and   be a subgroup of  . Then right and left cosets of   have the same number of elements.

Proof: Since   is a left and a right coset we immediately have   for all  .

Corollary 14: Let   be a group and   be a subgroup of  . Then the number of left cosets of   in   and the number of right cosets of   in   are equal.

Proof: By Lagrange's theorem and its right coset counterpart, we have  . We immediately obtain  , as was to be shown.

Now that we have developed a reasonable body of theory, let us look at our first important family of groups, namely the cyclic groups.

ProblemsEdit

Problem 1 (Matrix groups): Show that:

i) The group   of invertible   matrices is a subgroup of  . This group is called the general linear group of order  .
ii) The group   of   orthogonal matrices is a subgroup of  . This group is called the orthogonal group of order  .
iii) The group   is a subgroup of  . This group is called the special orthogonal group of roder  .
iv) The group   of unitary matrices is a subgroup of  . This is called the unitary group of order  .
v) The group   is a subgroup of  . This is called the special unitary group of order  .

Problem 2: Show that if   are subgroups of  , then   is a subgroup of   if and only if   or  .