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June 4th, 2014, 10:03 PM   #1
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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.
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June 5th, 2014, 12:14 AM   #2
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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 $3\Bbb Z = \{k \in \Bbb Z: k = 3t, t\in \Bbb Z\}$?
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June 5th, 2014, 12:59 AM   #3
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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.

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Last edited by Kala; June 5th, 2014 at 01:16 AM.
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June 5th, 2014, 01:49 AM   #4
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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:

$\text{gcd}(a,b) = \max(\{d \in \Bbb Z^+: d|a \text{ and } d|b\})$
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June 5th, 2014, 03:36 AM   #5
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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<r<d and r,q in Z
then r=a-dq =a-q(ax+by) = a(1-qx) +(-qy)b which contradicts as 0<r<d and d is the least element so d|a
same with d|b.

If d is not the largest number that divides both a and b then some other number exists. Let it be j such that j|ka +mb.
j|ka+mb=d
j|d
so j<=d
so d is the largest number that satisfies the conditions.
so gcd(a,b)=max({d in Z+ : d|a and d|b})


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June 6th, 2014, 09:49 AM   #6
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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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