Algebra

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  G1G2  ≠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  G
∩ 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

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