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Help Proving Isomorphism of a group

Bruce Wayne

Member
Apr 6, 2013
16
Hi!

I'm trying to prove a cyclic group is isomorphic to ring under addition. What the strategy I would take? How would I get it started?

Here's what I know so far:

I need to meet 3 conditions-- 1 to 1, onto, and the operation is preserved. I also know that isomorphic means that the group is homomorphic with onto and 1 to 1.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
I'm trying to prove a cyclic group is isomorphic to ring under addition. What the strategy I would take? How would I get it started?
You need to come up with an explicit definition of an alleged isomorphism.
 

Poirot

Banned
Feb 15, 2012
250
are you tring to prove a cyclic group G of order n is isomorphic to Z_n ={0,.......,n-1}?


Define $f(g)=g^n$ (mod n)
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
are you tring to prove a cyclic group G of order n is isomorphic to Z_n ={0,.......,n-1}?


Define $f(g)=g^n$ (mod n)
I may be missing something. Is $f:G\to\mathbb{Z}_n$? Then what is $g^n\pmod{n}$?
 

Bruce Wayne

Member
Apr 6, 2013
16
are you tring to prove a cyclic group G of order n is isomorphic to Z_n ={0,.......,n-1}?


Define $f(g)=g^n$ (mod n)
Yes, exactly. I want to show that a cyclic group G of order n is isomorphic to Zn.

I understand the concepts, and I know how to prove it for a relatively small n, but I haven't been able to find a completed proof online. I'd like to read it, and ask a few questions on the theory of why it works (should any arise).
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Since $G$ is cyclic of order $n$, there exists a $g\in G$ such that every $x\in G$ can be represented as $g^k$ for some $0\le k<n$. Therefore, we can define an $f:G\to\mathbb{Z}_n$ as follows: $f(g^k)=k$. It rests to show that $f$ is one-to-one, onto and a homomorphism.
 

Poirot

Banned
Feb 15, 2012
250
I may be missing something. Is $f:G\to\mathbb{Z}_n$? Then what is $g^n\pmod{n}$?
yes sorry I mean f(g^k)=k (mod n), where g is the generator of G