Algebra

   Questions Answers
 

Let S= {0,1,2,3,4,5.6.7} and  ⊗ denotes multiplication modulo 8 , i.e. x ⊗ y= (xy) mod 8

(i) Prove that ({0,1}, ⊗ } is not a group
(ii) Write 3 distinct groups (G,⊗) where G S and  G has 2 elements
 

Solution 
(i) ⊗ does not have an inverse
(ii) {1,3}, {1,5}, {1,7}

Let G₁ and G₂ be any two subgroups of G.
(i) Show that G₁ G₂ is also a subgroup of G
(ii) Is G₁ G₂  always a subgroup of G ?

Solution 
Let G₁ and G₂ be any two subgroups of G. Then  G₁∩G₂  ≠Φ , since at least identity element e is common to both G₁  and G₂ . the order to prove that G₁ ∩ G₂ is  subgroup it is sufficient to prove that

a∈G₁  ∩ G₂ , b ∈G₁ ∩ G₂

⇒ ab^-1∈G₁ ∩ G₂

Now, a∈G₁ ∩ G₂,    a ∈G₁ and a ∈G₂

 b∈G₁ ∩ G₂,    b ∈G₁ and b∈G₂

But G₁, G₂ are subgroups , therefore 
a ∈G₁, b∈G₁

⇒ ab^-1∈ G₁, a∈G₂, b ∈G₂

⇒ ab^-1 ∈ G₂
Finally ab^-1 ∈ G₁, ab^-1 ∈ G₂
⇒ ab^-1 ∈ G₁ ∩ G₂

Thus we have shown that a ∈ G₁ ∩ G₂ , b ∈ G₁ ∩ G₂ 

⇒ ab^-1 ∈ G₁ ∩G₂

Hence G₁ ∩G₂ is a subgroup of G

(ii) Solution : No

It is not necessarily that G₁ ∪ G₂ is a subgroup of G .
A group (G,*) must satisfy four conditions :
         closure,associtivity, identity and inverse
Since G₁ and G₂ are subgroups , ie groups themselves, G₁ ∩ G₂ also satisfy closure,associative and identity.
Since every element has its unique  G₁ ∩ G₂ is a subgroup ie if a has a^-1 in G , then a^-1 is also a inverse of a in G₁ and G₂
a^-1  ∈ G1 ∩ G2

Let (A,*) be a semigroup. Furthermore,for every a and b in A, if a   ≠ b, then a* b  ≠ b*a
(i)  Show that for every a in A , a*a=a
(ii) Show that for every a,b in A , a* b*a =a
(iii) Show that for every a,b,c in A , a*b*c = a*c

Solution
(i) A is a semigroup
∴ (a*b) * c = a*(b*c)
Putting, b=a  and c=a , we get
      (a*a)*a = a * (a*a) and A is not abelian
     a*a =a
⇒ b*b=b
(ii) Let b ∈ A , then 
Premultiplying by a, we get 
                  a*(b*b) =a*b
                  (a*b)*b=(a*b)
 ⇒               a*b = a
 ⇒               a*b*a = (a*b)*a
 

Using Equation (ii) , a*b*a=a*a
            =a 
(III) a*b*c=(a*b)*c
             = a*c
 

Let ({p*q},*) be a semigroup where p*p =q. Show that
(i) p*q=q*p
(ii) q*q=q

Solution
Let S = ({p,q},*) be a semi group
Since S is a semi group, it is closed and associative, ie
a) a*b ∈ S a v b ∈ S
b) a*(b*c)=(a*b)*c∀ a,b,c ∈ S
(i)                  p*q = p*(p*p)
                           = (p*p)*p
                          = q*p
(ii)  to prove q*q =q
Since , S is closed p*q ∈ S = {p,q}
therefore  p*q =p, or p*q=q
Case I - Let p*q =p
               q*q =q*(p*p)
                    = (q*p)*p
                   = (p*q)*p
                   =p*p
                 = q
Case II: Let p*q=q 
                 q*q =(p*p)*q
                      = p*(p*q)
                      = (p*q)
                      =p*q 
                      = q