Conventional Operators in Group Theory Homework

Click For Summary

Homework Help Overview

The discussion revolves around the topic of group theory, specifically concerning the conventional operators used in group definitions. The original poster questions whether a cyclic group of order 14 is isomorphic to Z mod 14 under addition or multiplication, noting that the isomorphism holds under addition but not under multiplication.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to clarify the appropriate operator for Z mod 14, questioning if it should be assumed to be addition due to the properties of cyclic groups.

Discussion Status

Participants have provided insights regarding the nature of Z mod 14 under different operations, noting that it is not a group under multiplication due to the lack of inverses for all elements. This has led to a consensus that addition is the conventional operator in this context.

Contextual Notes

There is an ongoing discussion about the implications of using different operators and the definitions of groups, particularly regarding the properties of elements and their inverses in Z mod 14.

chinye11
Messages
22
Reaction score
0

Homework Statement


I've just started to study group theory, and i keep encountering questions where no operators are specified so i was wondering if there was a conventional operator that was meant to be used. For instance I had a question to prove that a cyclic group of order 14 is isomorphic with Z mod 14. This is true under addition but not under multiplication so should i presume that Z14 with unspecified operator is addition?


Homework Equations


None.


The Attempt at a Solution

 
Physics news on Phys.org
Welcome to PF, chinye11! :smile:

Z/14Z is not a group under multiplication.
(Why not?)

So it has to be addition.
As a bonus it is isomorphic to C14 (also written as Z14).
 
Z/14 is not a group under multiplication because not every element has an inverse, my bad didn't check it.
 
chinye11 said:
Z/14 is not a group under multiplication because not every element has an inverse, my bad didn't check it.

Yep!

The group Z mod 14 with multiplication is denoted as (Z/14Z)x or as (Z/14Z)*.
That is, the same set, but with all elements that do not have an inverse removed.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K