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__Unit-1|Elementary Group Theory__**1.(a) If a * b = 3a+2b for a, b ∈ z, the set of integers, show that * is a binary operation on z.**

**Solution:**

Given: a * b =3a+2b for a, b ∈ z

Closure property

If a,b ∈ z then a+b ∈ z eg.If a=1 and b=2,thena * b=3a+2b=3⨯1+2⨯2=3+4=7 ∈ z. It is unique.∴ * is a binary operation on z.

**(b) Define Abelian group**

**Abelian group**:A group (G,*) is said to be an abelian group if a * b = b * a for all a,b ∈ G.

**Permutation group:**Let S be finite and has n elements, the group A (S) is also denoted by Sₙ. It has n! elements and is also called group of permutations of degree n or symmetric group of degree n.

**(2) Given the algebraic structure (G, *) with G={1,ω,ω²} where ω represents the cube root of unity and * stands for the binary operation of ordinary multiplication of complex numbers. Show that (G, *) is a group.**

**Solution:**

The Cayley's table is given below:

x | 1 | ω | ω² |

1 | 1 | ω | ω² |

ω | ω | ω² | 1 |

ω² | ω² | 1 | ω |

(i) Since 1×1= 1, 1 × ω=ω×1=ω and 1×ω²=ω²×1=ω²

So, 1 is the multiplicative identity.

(ii) Since 1×1=1, so, 1is the multiplicative inverse of itself.

(iii)Since ω×ω²=ω²×ω=1

So, ω is the inverse of ω² ans ω² is the inverse of ω.

(iv) ω×ω²=1∈G and 1 × ω= ω∈G .It shows that the product of any two elements of G is also an element of G. So the operation * is closed.

(v) (1×ω)×ω²=ω×ω²=1 and 1 ×(ω×ω²)=1×1=1

∴(1×ω)×ω²=1×(ω×ω²)=1

∴ It also satisfies the associative, property

Thus,(G, *) is a group.

The composition tables are given below:

So, 1 is the multiplicative identity.

(ii) Since 1×1=1, so, 1is the multiplicative inverse of itself.

(iii)Since ω×ω²=ω²×ω=1

So, ω is the inverse of ω² ans ω² is the inverse of ω.

(iv) ω×ω²=1∈G and 1 × ω= ω∈G .It shows that the product of any two elements of G is also an element of G. So the operation * is closed.

(v) (1×ω)×ω²=ω×ω²=1 and 1 ×(ω×ω²)=1×1=1

∴(1×ω)×ω²=1×(ω×ω²)=1

∴ It also satisfies the associative, property

Thus,(G, *) is a group.

**(3) Prepare the composition table of the addition modulo 3 and the multiplication modulo 3 for the set G={0,1,2}****Solution:**The composition tables are given below:

+₃ | 0 | 1 | 2 |

0 | 0 | 1 | 2 |

1 | 1 | 2 | 0 |

2 | 2 | 0 | 1 |

x₃ | 0 | 1 | 2 |

0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 |

2 | 0 | 2 | 1 |

**(4).Show that the set of integers Z forms a group under the operation of addition.**

**Solution:**

a+b ∈ Z and is unique

∴Z is closed under addition

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

∴It holds associative law.

0∈G is identity,Since 0+a = a=a+0,∀a ∈ G

For any a ∈ G,we have -a ∈ G and

a+(-a)=0=(-a)+a

This all shows that G is a group under addition.

**(5) If a,b∈ (G, *), then (i) (a⁻¹)⁻¹ =a, (ii) (a, b)⁻¹=b⁻¹a⁻¹**

**Proof :**(i) We have a⁻¹ * a=e

Multiplying both sides on the left by (a⁻¹)⁻¹,we get

(a⁻¹)⁻¹ * (a⁻¹ * a)= (a⁻¹)⁻¹ * e

or, [(a⁻¹)⁻¹ * a⁻¹] * a =(a⁻¹)⁻¹ (by associativity law) e * a= (a⁻¹)⁻¹=a=(a⁻¹)¹

(ii) We have, (a * b) (b⁻¹ * a⁻¹) = a * (b * b⁻¹) *a⁻¹(by Associative law)

=a * e * a⁻¹ (by defⁿ of inverse) a * a⁻¹ (by defⁿ of identity)

=e (by defⁿ of inverse) and similarly (b⁻¹ * a⁻¹) (a * b)

=e. So, b⁻¹ * a⁻¹ is the inverse of a * b, i.e. (a,b)⁻¹= b⁻¹ * a⁻¹

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**6) Show that the multiplication is a binary operation on the set S={-1,0,1} but the subtraction is not.**

**Solution:**

Let us consider the operation of multiplication.

Let x,y,∈ S or not ∀ x,y ∈ S.

Since,-1 * 0= 0 ∈ S, 0 * 1 = 0 ∈ S

1 *(-1)= -1 ∈ S, 1* 1 = 1 ∈ S and -1 * (-1) =1∈ S, 0 * 0= 0

So, multiplication is the binary operation.

Again,the operation '-' , Let x-y ∈ S or not for all x,y ∈ S.

Since, -1-0=-1∈ S,1-0=1 ∈ S

but-1-1 =2 doesn't belongs to S•

∴ Subtraction is not the binary operation•

**(7) Given a set G ={ 0,1} and a binary operation + defined by :**

**0 +0 =0; 0+1 = 1**

**1+0= 1; 1+ 1 =0**

**Find the additive identity and additive inverse of 0 and 1.**

+ | 0 | 1 | |

0 | 0 | 1 | |

1 | 1 | 0 | |

**Solution:**

(a) Since 0+0=0; 0+1 =1+0, so 0 is the additive identity,

(b) Since 0+0= 0; 1+1 =0, so, 0 is the additive inverse of 0, and 1 is the additive inverse of 1. In other words , every element of G is the additive inverse of itself.