Determine, in each of the following cases, whether the described system is or not a group. explain your answers. determine what is an abelian group.
a) g = {set of integers , a ∗ b = a − b}
b) g = {set of matrices of size 2 × 2, a ∗ b = a · b}
c) g = {a0, a1, a2, a3, a4ai ∗aj = aii+ji , if i+j < 5, ai ∗aj = aii+j-5i , if i+j ≥ 5}

(a) Not a group

(b) Not a group

(c) Abelian group

Step-by-step explanation:

In order for a system <G,*> to be a group, the following must be satisfied

(1) The binary operation is associative, i.e., (a*b)*c = a*(b*c) for all a,b,c in G

(2) There is an identity element, i.e., there is an element e such that a*e = e*a = a for all a in G

(3) For each a in G, there is an inverse, i.e, another element a' in G such that a*a' = a'*a = e (the identity)

If in addition the operation * is commutative (a*b = b*a for every a,b in G), then the group is said to be Abelian

(a)  

The system <G,*> is not a group since there are no identity.  

To see this, suppose there is an element e such that  

a*e = a

then  

a-e = a which implies e=0

It is easy to see that 0 cannot be an identity.

For example  

2*0 = 2-0 = 2

Whereas

0*2 = 0-2 = -2

So 2*0 is not equal to 0*2

(b)

The system <G,*> is not a group either.

If A is a matrix 2x2 and the determinant of A det(A)=0, then the inverse of A does not exist.

(c)

The table of the operation G is showed in the attachment.

It is evident that this system is isomorphic under the identity map, to the cyclic group

the system formed by the subset of Z, {0,1,2,3,4} with the operation of addition module 5, which is an Abelian cyclic group

We conclude that the system <G,*> is Abelian.

Attachment: Table for the operation * in (c)

the answer is true.

step-by-step explanation:

don't know what that says sorry i can't you

step-by-step explanation:

idk what it says