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)

answer: the first graph

step-by-step explanation:

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!