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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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    Are you happy with why the subgroup has the form $d \mathbb{Z} $?2011-11-15

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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.2011-11-14
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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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    @Andrew: Just glad you liked it. I figured a little explanation and generalization here and there would help out :D2011-11-15