Thread: Abstract Math Group

1) Find all the distinct generators of the group w.r.t. multiplication.

2) Find all the subgroups of group


Is there a way to find all the distinct generators without computing every element in ?

Hi

Of course there is a way: let be an integer, the generators of are the classes of integers which are relatively prime with i.e.

That comes from Bezout relation: so is a generator.

Conversely, generator

Originally Posted by clic-clac

Hi

Of course there is a way: let be an integer, the generators of are the classes of integers which are relatively prime with i.e.

That comes from Bezout relation: so is a generator.

Conversely, generator

Does the GCD=1 work for the group with respect to multiplication?

Because since 11 is prime, every number less than 11 must also be relatively prime, but not all those numbers are distinct generators. Those are [2],[6],[7],[8]. Can you use some trick to figure those out without testing all the elements in ?

Multiplication? I don't understand, is an additive group, and since is a prime, every non-zero class is a generator.

Originally Posted by clic-clac

Multiplication? I don't understand, is an additive group, and since is a prime, every non-zero class is a generator.

can be a multiplicative group also. All the non-zero elements of ([1],[2],[3],[4].....[11]) also form a group under multiplication. This can be verified using a Cayley table. In fact, for any when n is prime, it forms a group w.r.t. multiplication.

It can easily be verified that [2],[6],[7],[8] are distinct generators for the group under multiplication. But I was wondering if there's an easier way to get those answers, like the ones for the additive group just by knowing which element is relatively prime to n?


EDIT:
For multiplication, for example [10] only generate the set {[1],[10} and [3] generates the set {[1],[3],[9],[5],[4]} and etc. But the classes I mentioned above will generate the whole group. I got those values by doing it manually, but if n is large like , it will be a very laborious process.

Originally Posted by TitaniumX

([1],[2],[3],[4].....[11])

Is the last [10]?

If yes ok, I see that's what I call
This group is isomorphic to

So it's a cyclic group, and since only 1,3,7,9 are relative primes with 10, it has 4 distinct generators.

To easily find the generators, maybe there are formulas but I don't see one now.
What can be done is, if you know a generator of , for instance then the group isomomorphism \mathbb{Z}_{10},+)\rightarrow (\mathbb{Z}_{11}^*,.)" alt="\psi\mathbb{Z}_{10},+)\rightarrow (\mathbb{Z}_{11}^*,.)" /> such that gives you the others generators:
Indeed, and