# Abstract Algebra – Group, Subgroup, Abelian group, Cyclic group

Here in this post we will discuss about group, subgroup, abelian group, cyclic group and their properties.

## Group

A non-empty set *G* is said to form a group with respect to an operation *o*, if

G is closed under the operation i.e. a *o* b ∈ *G* for all a, b ∈ *G [groupoid]*

*o*is associative i.e. a

*o*(b

*o*c) = (a

*o*b)

*o*c for all a, b, c ∈

*G*

**[semigroup]**There exists an identity element

*e*in

*G*such that a

*o e*= a for all a ∈

*G*

**[monoid]**For each element a in

*G*, there exists an inverse element a

^{-1}in

*G*such that a

*o*a

^{-1}=

*e*

Example |

The set of all n x n matrices under the operation of matrix multiplication is not a group since not every n x n matrix has its multiplicative inverse but if *G* is the set of all n x n non-singular matrices, then *G* forms a group under the operation of matrix multiplication.

## Some elementary properties of groups

1. The identity element *e* is unique in a group (*G*, *o*).

2. The inverse a^{-1} of the element a ∈ *G* is unique.

3. For every a ∈ *G*, (a^{-1})^{-1} = a.

4. For all a, b ∈ *G*, (a *o* b)^{-1} = b^{-1} *o* a^{-1}.

5. The cancellation laws

a *o* b = a *o* c ⇒ b = c [left cancellation law]

b *o* a = c *o* a ⇒ b = c [right cancellation law] hold in *G*; for all a, b, c ∈ *G*

6. For a, b ∈ *G*, the linear equation a *o* x = b and y *o* a = b have unique solution in *G*

## Abelian group

A group (*G*, *o*) is said to be abelian or commutative if *o* is commutative i.e. a *o* b = b *o* a for all a, b ∈ *G*.

Example |

**Show that the set of all rational numbers is an abelian group with respect to addition.**

**Solution:**

Let us consider the set of all rational numbers *Q* with the binary operation addition. Then for any a, b, c ∈ *Q* we have,

i) a + b ∈ *Q*, as the sum of rational numbers is rational.

ii) a + (b + c) = (a + b) +c

iii) a + 0 = 0 + a = a for all a ∈ *Q*, so 0 ∈ *Q* is the identity element.

iv) a + (-a) = (-a) + a = 0 for all ∈ *Q* so the inverse element (-a) exists for each element a ∈ *Q.*

Hence the set of all rational numbers form a group with respect to addition, i.e. (*Q*, +) is a group.

Again a + b = b + a for all a, b ∈ *Q* so (*Q*, +) is an abelian group.

## Order of group

The order of an element *a* in a group *G* is the smallest positive integer n such that a^{n} = *e*, the identity element in *G*. Here, by an we understand a *o* a *o* … *o* a (n factors). If there is no finite integer n such that a^{n} = *e*, we say that the element a has order 0. If an element a ∈ *G* has order n, we write *O*(a) = n.

The order of a group is the number of elements in the group. Thus, if a group *G* has n elements, then *G* is said to be **finite group **of order n, and we write *O*(*G*) = n.

Example |

**If G is a group of even order, prove that it has an element a ≠ e satisfying a^{2} = e.**

**Solution:**

Suppose that *G* is a group of order 2n, n being a positive integer. Let a_{1}, a_{2}, …, a_{2n – 1}, e be the elements of *G*. Since in a group every element has its unique inverse and since there are odd number of elements a_{1}, a_{2}, …, a_{2n – 1} none of which is the identity element of *G*, it follows that there is one element a (say) among a_{1}, a_{2}, …, a_{2n – 1} whose inverse is a itself. Then it follows that a o a =e i.e. a^{2} = *e*, a ∈ *G* and a ≠ *e*.

Example |

**Show that if every element of the group G is its own inverse, then G is abelian.**

**Solution:**

Let a, b ∈ *G*, then a *o* b ∈ *G*. From the given condition a^{-1} = a, b^{-1} = b, (a *o* b)^{-1} = (a *o* b)

But (a *o* b)^{-1} = b^{-1} *o* a^{-1}, so that a *o* b = b^{-1} *o* a^{-1}

Or, a *o* b = b *o* a.

Hence *G* is abelian.

## Subgroup

Let (*G*, *o*) be a group and *H* be a non-empty subset of *G*. If (*H*, *o*) is a group where o is the induced composition, the (*H*, *o*) is said to be a subgroup of (*G*, *o*).

Example |

**( R, +) is a group. Z is a non-empty subset of R and (Z, +) is a group. Therefore, (Z, +) is a subgroup of (R, +).**

Theorem |

**Let ( G, o) be a group. A non-empty subset H of G forms a subgroup of (G, o) if and only if a ∈ H, b ∈ H ⇒ a o b^{-1} ∈ H.**

**Proof:**

Let (*H*, *o*) be a subgroup of (*G*, *o*).

Since (*G*, *o*) is a group, b ∈ *H* ⇒ b^{-1} ∈ *H* and therefore a ∈ *H*, b ∈ *H* ⇒ a *o* b^{-1} ∈ *H*.

Conversely,

Let *H* be a non-empty subset of *G* such that a ∈ *H*, b ∈ *H* ⇒ a *o* b^{-1} ∈ *H*.

Let a ∈ *H *then a ∈ *H*, a ∈ *H* ⇒ a *o* a^{-1} = *e* ∈ *H*. Therefore *H* contains identity element.

Now *e* ∈ *H*, a ∈ *H* ⇒ *e* *o* a^{-1} = a^{-1} ∈ *H*. Hence a ∈ *H* ⇒ a^{-1} ∈ *H*. Therefore the inverse of each element in *H* exists in *H*.

Let a ∈ *H*, b ∈ *H* then a ∈ *H*, b^{-1} ∈ *H* and by the given condition a *o* (b^{-1})^{-1} = a *o* b ∈ *H*. Hence a ∈ *H*, b ∈ *H* ⇒ a *o* b ∈ *H*, therefore *H* is closed under *o*.

Since *H* is a non-empty subset of *G* and *o* is associative on *G*, *o* is associative on *H*.

Therefore, (*H*, *o*) is a group and hence (*H*, *o*) is a subgroup of (*G*, *o*).

## Finite group

A group (*G*, *o*) is said to be a finite group if *G* contains a finite number of elements.

## Cyclic group

A group (*G*, *o*) is said to be a cyclic group if there exists an element a in *G* such that *G* = {a^{n} : n ∈ *G*} i.e. *G* = *<a>,* a is said to be a generator of the cyclic group.

Example |

**( Z, +) is cyclic group generated by 1, -1.**

**Let**

*S*= {1,*i*, -1, –*i*}. Then (*S*, .) is a cyclic group generated by*i*, –*i*. Theorem |

**Every cyclic group is abelian.**

**Proof:**

Let (*G*, *o*) is a cyclic group, generated by a.

Let p, q ∈ *G* then p = a^{r}, q = a^{s} for some integer r and s.

p *o* q = a^{r} *o* a^{s} = a^{r + s}

q *o* p = a^{s} *o* a^{r }= a^{s + r}

Since r + s = s + r, p *o* q = q *o* p for all p, q ∈ *G*.

Therefore the group is abelian.

## Left Coset

Let *G* be a group and *H* be subgroup of *G*. Let a be an element of *G* for all h ∈ *H*, ah ∈ *G*. The subset { ah : h ∈ *H* } is called a left coset of *H* in *G* and is denoted as *aH*.

## Right Coset

Let *G* be a group and *H* be subgroup of *G*. Let a be an element of *G* for all h ∈ *H*, ha ∈ *G*. The subset { ha : h ∈ *H* } is called a right coset of *H* in *G* and is denoted as *Ha*.

It may easily be shown that any two right (or left) cosets of *H* in *G* are either identical or have no element in common.

Theorem |

**(Lagrange) If G is a finite group and H is subgroup of G, then O(H) is a divisor of O(G).**

Theorem |

**Order of a cyclic group is equal to the order of its generator.**

Theorem |

**A subgroup of a cyclic group is cyclic.**

What’s Happening i’m new to this, I stumbled upon this I’ve found It positively useful and it has helped me out loads. I am hoping to give a contribution & aid different customers like its helped me. Good job.