Question

1. Let G = {a, b, c, d, e} be a set with an associative binary operation multiplication such
that ab = ba = d, ed = de = c. Prove that G under this multiplication cannot consist
of a group.
Hint: Assume that G under this operation does consist of a group. Try to complete
the multiplication table and deduce a contradiction.

2. Let G be a group containing 4 elements a, b, c, and d. Under the group operation called
the multiplication, we know that ab = d and c2 = d. Which element is b2? How about
bc? Justify your answer.

Answer 1

1. We have G={a,b,c,d,e} where ab=ba=d and ed=de=c.

A group G can exist under set G and binary operation 'multiplication' if:

• It conatins an identity.
• It contains an inverse.
• The operation is associative.
• And the group is closed under the operation.

Let us assume that G has an element x which is an identity element, which implies:

• a*x=a
• b*x=b
• c*x=c
• d*x=d
• e*x=e, this means that one of the values of set G has to be 1 for there to exist an identity element.

Now, let us assume that G has an element y which is an identity element, which implies:

• a*y=1
• b*y=1
• c*y=1
• d*y=1
• e*y=1, this gives multiple values to y, 1/a,1/b,1/c,1/d and 1/e which is not possible and hence

G under this multiplication cannot consist of a group.

2. G={a,b,c,d} and ab=d, c2 =d and also, G is a group which tells us that it has an identity element, an inverser, it is associative and closed under the multiplication operation.

Substituting the given values, we have G={a,b,(ab)1/2 ,ab}

Keeping in mind the properties of a group, there exists an identity and an inverse. Looking at available vallues, we can clearly see that (ab)1/2 will need to have an inverse too, which will only be possible if it was 1, which implies ab is 1, thus making a and b the inverses of each other.

So, a=1/b which also implies b=1/a, and c=1, d=1.

So, b2 =1/a2 and

bc=1/a*1=1/a.