What operator is assumed for Z14 with no specification?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
3 replies · 2K views
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

 

Discussion

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.