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This wonderful restatement was taken from Keith Conrad's notes: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cyclicgp.pdf?fbclid=IwAR2Tve3QPGsIEZAfj5sVhoKDdhXT2oPMsSnGoKyhZGc1NxFcynmk1_uJNRE

Theorem: $\forall a,b\Bbb{Z}$, we have $a\Bbb{Z}+b\Bbb{Z}=\gcd(a,b)\Bbb{Z}$.

Proof attempt: If $\gcd(a,b)=d$, then $a=\alpha\cdot d$ and $b=\beta\cdot d$ such that $\gcd(\alpha,\beta)=1$. Thus, $a\Bbb{Z}+b\Bbb{Z}=\alpha\cdot d\Bbb{Z}+ \beta\cdot d\Bbb{Z}=(\alpha+\beta)d\Bbb{Z}$. Since $(\alpha +\beta)\in\Bbb{Z}$ implies $(\alpha+\beta)d\in d\Bbb{Z}$. By properties of cosets, $(\alpha+\beta)d\Bbb{Z}=d\Bbb{Z}$.

TheLast Cipher
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  • The simple proof is already fully presented there (Theorem A.1 p. 8). Did you not understand it? This is a standard result that will become clearer when you study ideal theory of rings. – Bill Dubuque Nov 08 '18 at 04:15
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    Remark the Conrad's claim that "The proof here is completely independent of Euclid’s algorithm" is a bit misleading in my opinion. The proof he gives uses the fact that subgroups of a cyclic group are cyclic, whose proof uses the (Euclidean) Division Algorithm - which is precisely the inductuve / descent step in Euclid's algorithm. It's the group-theoretic form of the proof I give here in more elementary language. – Bill Dubuque Nov 08 '18 at 04:24
  • @BillDubuque:I did not understand his proof. I don't see why $c$ being a common divisor of $a$ and $b$ implies $c\in a\Bbb{Z}+b\Bbb{Z}$. I also find it amusing that using the fact subgroups of a cylic group is cyclic makes the proof cylic. – TheLast Cipher Nov 08 '18 at 04:37
  • I think I get it now. nevermind. lol. – TheLast Cipher Nov 08 '18 at 04:38

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