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"Let $f: \mathbb{Z}^2 \to \mathbb{Z}^2$ be a group homomorphism and suppose that $f(a,b) = (c,d)$. Prove that: $\gcd(a,b) \mid \gcd(c,d)$. Prove that $\gcd(a,b) = \gcd(c,d)$ if $f$ is an automorphism." I am seeking for some hint to start this problem. Although I generally have no trouble with group theoretic notions, the smallest introduction of number theory confuses me terribly. Any suggestions would be greatly appreciated. |
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We have $\gcd(a,b)|c$ and $\gcd(a,b)|d$ because $$(c,d)=f(a,b)=\gcd(a,b)\cdot f\left(\frac a{\gcd(a,b)},\frac b{\gcd(a,b)}\right)$$ For an automorphism, there is an invers $g$ such that $g(c,d)=(a,b)$, hence you have both $\gcd(a,b)|\gcd(c,d)$ and $\gcd(c,d)|\gcd(a,b)$. Since both gcds are positive integers, equality follows. |
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Hint: Let $m = gcd(a,b)$. Then $a = me, b = mg$ for some $e , g \in \mathbb{Z}$. Since $f$ is a group homomorphism, we have $(c,d) = f(a,b) = f(m(e,g)) = mf(e,g) = m(h,j) = (mh,mj)$, where $f(e,g) = (h,j)$. Thus, $m|c,d$. What can you conclude from this? Also, for the automorphism case, using $f^{-1}$ and applying the previous case again, you get $gcd(c,d) | gcd(a,b)$. Since $gcd(c,d), gcd(a,b)$ are both non-negative, what can you conclude? |
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