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