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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.

Thanks!

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Start with the definition of homomorphism. What does this definition allow you to say about what the homomorphisms could and could not be? –  Arkamis Nov 27 '12 at 22:33
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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

1 Answer 1

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?

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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
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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

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