1. ## Abstract Math Group

1) Find all the distinct generators of the group $\mathbb{Z}_{11}$ w.r.t. multiplication.

2) Find all the subgroups of group $\mathbb{Z}_{11}$

Is there a way to find all the distinct generators without computing every element in $\mathbb{Z}_{11}$?

2. Hi

Of course there is a way: let $n$ be an integer, the generators of $\mathbb{Z}_n$ are the classes of integers which are relatively prime with $n,$ i.e. $\{[x]\in\mathbb{Z}_n;\ gcd(x,n)=1\}$

That comes from Bezout relation: $gcd(x,n)=1\Rightarrow \exists u,v\in\mathbb{Z},\ ux+vn=1 \Rightarrow [u][x]=[1]$ so $[x]$ is a generator.

Conversely, $[x]$ generator $\Rightarrow \exists u\in\mathbb{N},\ [u][x]=1\Rightarrow \exists u,v\in \mathbb{Z},\ ux=1-vn$

3. Originally Posted by clic-clac
Hi

Of course there is a way: let $n$ be an integer, the generators of $\mathbb{Z}_n$ are the classes of integers which are relatively prime with $n,$ i.e. $\{[x]\in\mathbb{Z}_n;\ gcd(x,n)=1\}$

That comes from Bezout relation: $gcd(x,n)=1\Rightarrow \exists u,v\in\mathbb{Z},\ ux+vn=1 \Rightarrow [u][x]=[1]$ so $[x]$ is a generator.

Conversely, $[x]$ generator $\Rightarrow \exists u\in\mathbb{N},\ [u][x]=1\Rightarrow \exists u,v\in \mathbb{Z},\ ux=1-vn$
Does the GCD=1 work for the group ${Z}_{11}$ 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 ${Z}_{11}$?

4. Multiplication? I don't understand, $\mathbb{Z}_{11}$ is an additive group, and since $11$ is a prime, every non-zero class is a generator.

5. Originally Posted by clic-clac
Multiplication? I don't understand, $\mathbb{Z}_{11}$ is an additive group, and since $11$ is a prime, every non-zero class is a generator.
${Z}_{11}$ can be a multiplicative group also. All the non-zero elements of ${Z}_{11}$ ([1],[2],[3],[4].....[11]) also form a group under multiplication. This can be verified using a Cayley table. In fact, for any ${Z}_{n}$ 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 ${Z}_{97}$, it will be a very laborious process.

6. Originally Posted by TitaniumX
([1],[2],[3],[4].....[11])
Is the last [10]?

If yes ok, I see that's what I call $\mathbb{Z}_{11}^*$
This group is isomorphic to $(\mathbb{Z}_{\phi(11)},+)=(\mathbb{Z}_{10},+)$

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 $\mathbb{Z}_{11}^*$, for instance $[2],$ then the group isomomorphism $\psi\mathbb{Z}_{10},+)\rightarrow (\mathbb{Z}_{11}^*,.)" alt="\psi\mathbb{Z}_{10},+)\rightarrow (\mathbb{Z}_{11}^*,.)" /> such that $[1]\mapsto [2]$ gives you the others generators: $\psi([3]),\ \psi([7]),\ \psi([9])$
Indeed, $\psi([3])=[2]^3=[8],\ \psi([7])=[8]^2[2]=[-3]^2[2]=[7]$ and $\psi([9])=[8]^3=[-3]^3=[-27]=[6]$