My Math Forum elements of the subgroup <3,12> of (Z, +)

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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.

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