Let F 3 6 9 12 and multiplication module 15 Convince me th

Let F = {3, 6, 9, 12}, and = multiplication module 15. Convince me that (F, ) is a group by constructing the Caley\'s table. What is e in F? Find the inverse of each element of F.

Solution

Given that F = { 3 , 6 , 9 , 12 } and * is the operation defined for any a , b in F we have ( a*b ) mod 15 = remainder of ( a x b) /15

Now we construct cayley\'s table for the set F = { 3 , 6 , 9 , 12 }

Form the table it is clear that all the elements are the elements of F.

So , * is binary on F

for any a , b, c in F it is clear that a*(b*c) = (a*b)*c

So, * is associative

For any a in F it is clear that a*6 = 6*a =6

So, 6 is identity element in F . That is e = 6.

The inverse of 3 is 12

The inverse of 6 is 6

The inverse of 9 is 9

The inverse of 12 is 3

So that each element of F has inverse .

Thus , F = {3, 6, 9, 12} is a group under the operation = multiplication module 15.