please solve 1(a) and 1(b) In this problem, we show that any group of order 3 mu

Question

please solve 1(a) and 1(b)

In this problem, we show that any group of order 3 must be cyclic. Let G be a group of order 3, and let e. a, b be the three elements of G, where e is the identity of the group. Prove that b must be the inverse of a. [Hint: explain why ab must equal e.] Using part (a), prove that G must be cyclic. Consider the group G = {e, a, b, c, d, f, g, h} whose Cayley table is given below. What is the center of the group? What is |a|? What is |b|? (Please show in some way how you found these.) Find a cyclic subgroup H of G such that |H | = 4. Find a noncyclic subgroup K of G such that |K| = 4.

Explanation / Answer

a) since group element follow closer property
so
ab and ba should belongs to G
but in that case
ab should be equal to
a or b or e
if
ab=b than
a=e which is contradiction
similarly
ab=a is also not true
similarly for ba

ba=e
so
ab=e=ba
and
a*a-1=e=ab=a-1*a=ba

so

a-1=b

b)

since

a belongs to G

so

a^2 too

but

a*a^2=a^3=e=a^2*a

a^2=b

since G contain

e,a,b

or

e,a,a^2

so

group G can be generated by a

so G is cyclic

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