Each equivalence class is a power of [g]

[evinda]

Hello! :)

I have to find an equivalence class $[g] \in \mathbb{Z_{15}}^{*}$ so that each equivalence class $\in \mathbb{Z}^{*}_{15}$ is a power of $[g]$.

$\mathbb{Z}^{*}_{15}=\{[1],[2],[4],[7],[8],[11],[13],[14]\}$

I tried several powers of the above classes,and I think that there is no equivalence class $[g] \in \mathbb{Z_{15}}^{*}$ so that each equivalence class $\in \mathbb{Z}^{*}_{15}$ is a power of $[g]$.Is it actually like that or am I wrong?? (Thinking)

 

[johng1]

You're exactly right. The multiplicative group of the integers mod 15 is the direct product of the multiplicative groups of Z₃ and Z₅ ; i.e the direct product of a cyclic group of order 2 and one of order 4, definitely not cyclic.
See the Wikipedia article Multiplicative group of integers modulo n - Wikipedia, the free encyclopedia

 

[evinda]

You're exactly right. The multiplicative group of the integers mod 15 is the direct product of the multiplicative groups of Z₃ and Z₅ ; i.e the direct product of a cyclic group of order 2 and one of order 4, definitely not cyclic.
See the Wikipedia article Multiplicative group of integers modulo n - Wikipedia, the free encyclopedia

Thank you very much!

 

[Deveno]

Re: each equivalence class is a power of [g]

In fact:

$\langle [1]\rangle = \{[1]\}$

$\langle [2]\rangle = \{[1],[2],[4],[8]\} = \langle [8]\rangle$

$\langle [4]\rangle = \{[1],[4]\}$

$\langle [7]\rangle = \{[1],[7],[4],[13]\} = \langle [13]\rangle$

$\langle [11]\rangle = \{[1],[11]\}$

$\langle [14]\rangle = \{[1],[14]\}$

which shows that every element has order 1,2 or 4, and that no element has order 8.

(for $g > 7$ it is easier to compute $\langle[g]\rangle$ as $\langle[-(15-g)]\rangle$).