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Pretty easy question probably: How do you prove that $\phi(k)=(n-k)\mod{n}$ satisfies the homomorphism property for the binary structures $\langle\mathbb{Z}_n,+\rangle$ and $\langle\mathbb{Z}_n,+\rangle$?

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What have you tried? –  mixedmath Sep 20 '12 at 0:12
    
I'm trying to prove it's an isomorphism. What I have now is $\phi(j+k)=(n-(j+k)(\mod{n})=(n-j-k)(\mod{n})$. I don't know how the transition from this to $\phi(j)+\phi(k)$ would be justified. I've proved the 1-1 and onto parts. –  Mike 105 Sep 20 '12 at 0:25
    
I changed $<\mathbb{Z}_n,+>$ to $\langle\mathbb{Z}_n,+\rangle$. That is standard usage. –  Michael Hardy Sep 20 '12 at 1:41
    
For general groups, the map $x \mapsto x^{-1}$ is not a homomorphism but for abelian groups it is. This is what you have here, written additively. –  lhf Sep 20 '12 at 2:02
    
I'm new to LaTeX: How do you do those angle brackets? –  Mike 105 Sep 20 '12 at 3:33

1 Answer 1

up vote 1 down vote accepted

You need to show that for all $a$ and $b$, we have $\phi(a+b)=\phi(a)+\phi(b)$. So, to use number-theoretic terminology, you need to show that if $x\equiv -a\pmod{n}$ and $y\equiv -b\pmod{n}$, then $x+y\equiv -(a+b)\pmod{n}$.

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Yes, working backwards worked for me. Thanks. –  Mike 105 Sep 20 '12 at 0:52

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