Weird group isomorphism problem

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 3K views
y2kevin
Messages
6
Reaction score
0

Homework Statement


Show that the group Z/<(a,b)> is isomorphic to Z if gcd(a,b)=1. Find generators of Z/<(a,b)>.


2. Relevant information
Please note that the question is asking for Z/<(a,b)>, not ZxZ/<(a,b)>. I am having trouble understanding the meaning behind <(a,b)> as a subgroup of Z. My professor only gave the hint that <(a,b)> is cyclic and has an analogous case for ZxZ/<(a,b)> if gcd(a,b)=1.


The Attempt at a Solution


From my understanding, <(a,b)> should represent the set generated by a and b, ie, <(a,b)>={ n*a+m*b | n and m are integers }. However, for a and b s.t. gcd(a,b)=1, there are integers n and m such that n*a+m*b=1, in which case every element in Z can be generated from this linear combination.

This is clearly not the case since Z/<(a,b)> would be Z/Z={0}.

Does anyone have any idea what else <(a,b)> could be in this context

Thanks.
 
Physics news on Phys.org
Normally, the group generated by a and b is written <a,b>. Also, gcd(a,b) is sometimes written (a,b). So maybe Z/<(a,b)> is actually Z/<1>. But that doesn't help either. Where did you find this problem?
 
It came up during class following the problem ZxZ/<(7,37)>. Apparently these two problems share some similar characteristics.
 
Have you tried working out the problem with ZxZ instead of Z? I'm pretty sure it should be ZxZ.