Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am wondering if anyone could help me with a question I have. The question states: "Describe all homomorphisms from the group $\Bbb Z_4$ to the group $\Bbb Z_8$. "

I'm not sure where to start.


share|cite|improve this question
Start with the definition of homomorphism. What does this definition allow you to say about what the homomorphisms could and could not be? – Emily Nov 27 '12 at 22:33
A homomorphism is a function that transforms one group to another group. So in this case, I start with Z_4 and go to Z_8. I believe that means that there exists a function that maps Z_4 to Z_8. I also know that if such a function exists then it means that f(ab) = f(a)f(b). What elements in Z_4 are mapped to Z_8 though? This is where I am getting stuck. – RedPotatoe Nov 28 '12 at 1:40
Hey, the group operation of $\Bbb Z_4$ and $\Bbb Z_8$ is the addition, and not the multiplication!! So, you rather look for $f$ which satisfies $f(a+b)=f(a)+f(b)$. – Berci Nov 28 '12 at 8:51

Hint: Considering $\Bbb Z_4=\{0,1,2,3\}$ with $+$, we'll have that $\phi(1)$, the image of $1$, totally describes all the homomorphism $\phi$. What can $\phi(1)$ be?

share|cite|improve this answer
I believe that ϕ(1)=1 and ϕ(1)=4. – RedPotatoe Nov 28 '12 at 1:44
Well, if $\phi(1)=1$ then $0=\phi(0)=\phi(1+1+1+1)=4$ which is not true already in $\Bbb Z_8$. The $\phi$ with $\phi(1)=4$ is fine. There are two more possibilities. – Berci Nov 28 '12 at 8:48
Hi, I'm a bit confused on what you wrote above. What does " 0=ϕ(0)=ϕ(1+1+1+1)=4" mean? I got a hold of another book and I think I understand the original question but I'm not sure if I'm correct. I created an addition table, first by taking all elements in Z4. The second table I created was an addition table with elements of Z8. This table will help me see what elements are mapped to the identity element. Is that correct? In the table Z4, I have 0+1=1, 3+2=1, and in Z8 I have 0+1=1 and 4+1=1. Thank you for all the help you've been giving me. I truly appreciate it! – RedPotatoe Nov 28 '12 at 13:50
Yes, you seem to be a bit confused. Now the group operation is $+$, and hence the 'identity element' (which doesn't hurt anybody in the group by applying the group operation) is not the $1$, but it is the zero, alias, $0$. When I wrote $\phi(0)=\phi(1+1+1+1)$, I used that, in $\Bbb Z_4$ where $\phi$ maps from, we have $1+1+1+1=0$. Then, using that $\phi$ is homomorphism: i.e. is friendly with the group operation (that is $+$), we have $\phi(1+1+1+1)=\phi(1)+\phi(1)+\phi(1)+\phi(1)$. Then applied your assumptions that $\phi(1)=1$. – Berci Nov 28 '12 at 16:48
All in all, there are $4$ such homomorphisms, I didn't count one above ($3$ would be really odd, anyway:) – Berci Nov 28 '12 at 16:49

Your Answer


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.