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 June 4th, 2014, 10:03 PM #1 Newbie   Joined: May 2014 From: 2219 Posts: 13 Thanks: 2 elements of the subgroup <3,12> of (Z, +) Hi, I'm trying to do this question but im not sure if im doing it right. Find the elements of the subgroup <3,12> of (Z, +). Show that this subgroup is cyclic. i get; <3,12>={3x +12y : x,y in Z} ={0,3,6,9,12,15,...) gcd(3,12)=3, => <3,12> does not equal Z, => the subgroup is not cyclic. But the question say it is, so what am i doing wrong? Any ideas or help would be great!! Kala Last edited by greg1313; June 5th, 2014 at 04:21 PM. June 5th, 2014, 12:14 AM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 What is your understanding of what "cyclic" means? Your analysis so far is correct: let me ask you this- if $k = 3x + 12y$, can you find some integer $d$ such that $k = 3d$? If so, then this shows that $\langle 3,12\rangle \subseteq 3\Bbb Z$. Conversely, given $k = 3d$, it should be obvious that we can take $x = d, y = 0$, which shows that $3\Bbb Z \subseteq \langle 3,12\rangle$. What does this tell you? Can you find a generator for $3\Bbb Z = \{k \in \Bbb Z: k = 3t, t\in \Bbb Z\}$? Thanks from Kala June 5th, 2014, 12:59 AM #3 Newbie   Joined: May 2014 From: 2219 Posts: 13 Thanks: 2 Ok i think im getting it, so <3,12> = 3Z since 3x +12y =3d for d in Z. so the gcd tells you what multiples can be generated in Z? The generator of 3Z is 3, right? since <3>={3t : t in Z} oh so it is cyclic. Thanks Last edited by Kala; June 5th, 2014 at 01:16 AM. June 5th, 2014, 01:49 AM #4 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Yes, in fact, one way of DEFINING the gcd(a,b) of two integers is: $\min(\{ka+mb > 0: k,m \in \Bbb Z\})$ that is to say the gcd is the smallest positive $\Bbb Z$-linear combination of a and b. It is an instructive exercise to try to show that this is equivalent to the definition: $\text{gcd}(a,b) = \max(\{d \in \Bbb Z^+: d|a \text{ and } d|b\})$ June 5th, 2014, 03:36 AM #5 Newbie   Joined: May 2014 From: 2219 Posts: 13 Thanks: 2 Okay ill try, gcd(a,b)=min({ka + mb>0 : k,m in Z}) it has to have a least element as it is non-empty and is a subset of the natural numbers. Let the least element be d. If gcd(a,b)=d =>ax +by=d for x,y in Z which satisfies the conditions on gcd(a,b) If d does not divide a then a=dq +r for 0
 June 6th, 2014, 09:49 AM #6 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 That's a bit jumbled, but your idea is correct. So that proves definition 1 implies definition 2. Can you prove it the other way? It should be clear that if $d|a$ and $d|b$, that if we let: $M = \{ka+mb:k,m \in \Bbb Z\}$ that for ANY $x \in M$, we have $d|x$. So the "catch" is to show that $\text{gcd}(a,b) \in M$, that is we can actually write the gcd in that form. Again, the (generalized) Euclidean algorithm is your friend, here. Tags <3, <3, 12>, 12>, elements, subgroup Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mrtamborineman10 Math Books 3 August 22nd, 2011 11:06 PM liptonpc Applied Math 4 January 25th, 2010 10:08 AM envision Abstract Algebra 3 October 4th, 2009 10:37 PM envision Abstract Algebra 1 October 4th, 2009 03:24 AM liptonpc Abstract Algebra 1 December 31st, 1969 04:00 PM

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