Graphing Equations. answer quick please!
1 answer:
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Answer:
(a) Not a group
(b) Not a group
(c) Abelian group
Step-by-step explanation:
<em>In order for a system <G,*> to be a group, the following must be satisfied </em>
<em> (1) The binary operation is associative, i.e., (a*b)*c = a*(b*c) for all a,b,c in G </em>
<em>(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 </em>
<em> (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) </em>
<em> </em>
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)
