MHB Each equivalence class is a power of [g]

evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
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)
 
Physics news on Phys.org
johng said:
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! :rolleyes:
 
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 $\langle[-(15-g)]\rangle$).
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...

Similar threads

Replies
3
Views
435
Replies
6
Views
1K
Replies
1
Views
2K
Replies
3
Views
1K
Replies
15
Views
2K
Replies
3
Views
2K
Replies
7
Views
2K
Back
Top