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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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    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.2012-09-20
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    I changed $<\mathbb{Z}_n,+>$ to $\langle\mathbb{Z}_n,+\rangle$. That is standard usage.2012-09-20
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    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.2012-09-20
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    I'm new to LaTeX: How do you do those angle brackets?2012-09-20
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    `\langle \rangle`2012-09-20
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    @Mike105 : Your question was answered by Douglas S. Stones, but notice that you can find the answer by clicking on "edit" on your question and seeing what appears there.2012-09-20

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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.2012-09-20