# Abstract Algebra/Group Theory/Subgroup

## Subgroups[edit]

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 (why?). 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.

## Problems[edit]

**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 .