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I was watching a video of an algebra lecture on line. The substitute teacher was presenting a pf. of a lemma:

$$m\mathbb{Z} + n\mathbb{Z} = \mbox{gcd}(m,n)\mathbb{Z}\quad\mbox{where }m,n\in\mathbb{Z}$$

I am a bit puzzled by the development of the proof in that the first assertion is that the LHS is a subgroup and has the form $d\,\mathbb{Z}$. And then goes on to use this to complete the proof.

My question (probably naive) is how does one know at the onset that the LHS is a subgroup - after all, the lemma is intended to prove what that sum is, and knowing that, it can then be claimed that the sum is also a subgroup.

In this regard, there was also a problem I saw in "Dummit and Foote" which asks to prove that the intersection of two subgroups is a subgroup. So if this is applicable to my question, it seems it has to first be determined what the intersection of $m\mathbb{Z}$ and $n\mathbb{Z}$ are. Which then leads to the gcd of the RHS.

Thanks from a self-studier.

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The intersection of m$\mathbb{Z}$ and n$\mathbb{Z}$ is LCM(m,n)$\mathbb{Z}$. It is in fact a subgroup, but different from GCD(m,n)$\mathbb{Z}$. m$\mathbb{Z}$ + n$\mathbb{Z}$ is the set of all numbers of the form $am+bn$, so $4 \in 8\mathbb{Z} + 12\mathbb{Z}$ –  Ross Millikan Nov 14 '11 at 23:10
    
@RossThanks for your help. –  Andrew Nov 14 '11 at 23:28
    
Are you happy with why the subgroup has the form $d \mathbb{Z} $? –  Daniel Freedman Nov 15 '11 at 2:37
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2 Answers

up vote 3 down vote accepted

Do you believe that $m\mathbf{Z}$ and $n\mathbf{Z}$ are subgroups of $\mathbf{Z}$? In general, if $H, K$ are subgroups of an abelian group $G$ (written additively) then \[ H + K = \{x + y : x \in H,\, y \in K\} \] is also a subgroup of $G$. Just check the three axioms.

It would be a good exercise to check that the intersection $m\mathbf{Z} \cap n\mathbf{Z}$ is actually $\operatorname{lcm}(m, n)\mathbf{Z}$. This makes sense: to be in the intersection, you have to be a multiple of both $m$ and $n$. Good luck with your self-study!

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@DylanThanks for your kind response. –  Andrew Nov 14 '11 at 23:27
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For a subset $G'$ of any group $G$ to be a subgroup, three criterion must be met.

1) The identity element $e$ must be an element of $G'$. It is, in this case, because the element $m0+n0 = 0$ is the identity.

2) For any two elements $x, y \in G'$, we must have $xy\in G'$. This is also easy to check, by the following calculation, with $x = ma + nb$ and $y = ma' + nb'$, then:

$$ (ma + nb) + (ma' + nb') = m(a+a')+n(b+b') \in G' $$

This property is generally called "closed under multiplication", or "under addition" in this case.

3) For any element $x\in G'$, we must have $x^{-1}\in G'$. This one is also not too hard, since if $x = ma + nb$, then we have

$$ x + (m(-a)+n(-b)) = ma-ma+mb-mb = 0 $$

so we have $x^{-1} = (m(-a)+n(-b)) \in G'$

And that concludes the proof.

The intersection of the two subgroups would be the subgroup lcm$(m, n)\mathbb{Z}$, while the sum is $\gcd(m, n)\mathbb{Z}$.

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@ArthurThanks also for your really helpful answer. I tried to check both answers, but the system would only allow one. –  Andrew Nov 14 '11 at 23:27
    
@Andrew: that is correct, you can only accept one answer. You can upvote (click the up arrows) as many as you want. –  Ross Millikan Nov 14 '11 at 23:30
    
@RossThanks just registered so that I could. –  Andrew Nov 14 '11 at 23:39
    
@Andrew: Just glad you liked it. I figured a little explanation and generalization here and there would help out :D –  Arthur Nov 15 '11 at 4:25
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