Question

Abstract Algebra
Show that Z5  Z3 is a cyclic group, and list all of the generators for the
group.

Answer 1

Z5 x Z3

Answer 2

If it's cyclic it has one generator, right? So one way would be to find a generator g and show that any element of the group can be written as a power of g.

Answer 3

i didnt went to class when the teacher was teaching this subject and now i am having troubles understanding it.i searched this question on google and found the answer but since english is not my native language, that doesnt help me either.if i want you to explain it a little more, would you help?

Answer 4

Ok, so Z3 x Z5 is the group where addition is defined by

(a,b) + (c,d) = ((a+b) mod 3, (c+d) mod 5)

So, what's (1,3) + (1,2) ?

Sorry if this is obvious, I need to figure out what you know and what you dont

Answer 5

should be (2,0). and 0 is because of mod 5 right?

Answer 6

Yeah. Do you understand the definition of a generator for a group and cyclic group?

Answer 7

i know that when G = {\[a ^{n}\] : neZ } , a is the generator of G. but this is cyclic group right?

Answer 8

oh sorry about that.let me correct

Answer 9

G = { a^n : neZ }

Answer 10

Yeah, cyclic group is one generated by a single element. I'll just call any such element a generator. So now I'm looking for an element (a,b) so that

{(a,b) , (a,b) + (a,b), (a,b) + (a,b) + (a,b), ...} = G

Answer 11

So the idea would be to give some specific values a,b for this works. This proves that the group is cyclic.

Answer 12

Then think about all the elements that can do this. Hint: think about greatest common divisors.

Answer 13

then <1,0> is a generator of Z5xZ3 since <1,0>={(1,0),(2,0),(3,0),(4,0),(5,0)} right?

Answer 14

(1,1) is a member of the group too and we didn't get that.

Answer 15

<1,1> = {(1,1),(2,2),(3,0),(4,1),(5,2)} is this a generator too?

Answer 16

(1,1) is a generator, because <(1,1)> = G. You only listed the first 5 elements but (0,2) + (1,1) = (1,2) which we you didn't list before. Either way, if you go all the way you'll get all 15 elements of the group.

If you can use the Lagrange theorem you could stop at 6.