Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


I'm trying to solve a group theory question involving morphisms:
How many different morphisms do there exist from $ C_n $ to $ C_m $?

Am I correct in saying if $f$ is a morphism, then $f(0) = 0$ and $f(a+b) = (a+b)f(1)$?
If yes, where do I go from here? And if no, how do I start?
Thanks!

Also, as a follow up question there's:
How many automorphisms exist from $C_n$ onto itself?
Since automorphisms map generators to generators, I have to map $1$ to a generator, so I have $\phi(n)$ choices, with $\phi$ being Euler's totient function?

share|improve this question
    
Possible duplicate of math.stackexchange.com/questions/120861/… –  lhf Aug 7 '12 at 13:02
    
@lhf Not necessarily. That question deals with whether $\psi(1_n) = 1_m$ is a homomorphism, and allthough the two questions are clearly related, I wouldn't say it's a duplicate. –  Arthur Aug 7 '12 at 13:05

1 Answer 1

up vote 2 down vote accepted

You are right in your assumption (the phrase used by mathematicians is "A homomorphism is uniquely determined by its image on the generators"), and from there you just have to make sure that the image of the generators behave like they originally do; in this case the one generator $1 \in C_n$ fulfills $n\cdot 1 = 0$

By your assumtion, the homomorphism $\psi$ is completely determined by $\psi(1)$. So the question reads "Where can I send $1$?", and the only limiting factor here is that in $C_n$, $n\cdot 1 = 0$, so we must have $n\cdot \psi(1) = 0$ in $C_m$. Now the question reads "Which elements in $C_m$ has order a divisor of $n$?". This is when we leave algebra and enter number theory, where the same question sounds "Which integers $k$ fulfills $k\cdot n \equiv 0 \mod{m}$?" And the answer is any multiple of $\frac{m}{\gcd(m, n)}$, of which there are $\gcd(m, n)$, including $0$.

As to your follow-up, you can map $1$ to any element $j$ with $\gcd(n, j) = 1$, of which there are $\phi(n)$. If you map it to any other element, it will not generate a surjection, since $1$ would not be in the image, and thus not an isomorphism. So you are right.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.