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 $\langle\mathbb{Z}_n,+\rangle$?
Proof of Homomorphism Property for $(n - k)\mod n$
1
$\begingroup$
abstract-algebra
-
0I'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
-
0I changed $<\mathbb{Z}_n,+>$ to $\langle\mathbb{Z}_n,+\rangle$. That is standard usage. – 2012-09-20
-
0For 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
-
0I'm new to LaTeX: How do you do those angle brackets? – 2012-09-20
-
0`\langle \rangle` – 2012-09-20
-
0@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
1 Answers
1
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 $x+y\equiv -(a+b)\pmod{n}$.
-
0Yes, working backwards worked for me. Thanks. – 2012-09-20