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I am trying to classify, $\frac{\mathbb{Z} \times \mathbb{Z}}{\langle(n,n)\rangle}$, $n=1,2,3,\cdots, $ using fundamental theorem of finitely generated abelian groups.

Inspire by @user134824 in Classifying the factor group $(\mathbb{Z} \times \mathbb{Z})/\langle (2, 2) \rangle$, I construct $\varphi: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}_n$ by $(k,l) \mapsto (k-l,l)$. Then its kernel is $\langle(n,n)\rangle$ and from the 1st isomorphism theorem, $\mathbb{Z} \times \mathbb{Z} \simeq \mathbb{Z}_n \times \mathbb{Z}$.

But how I can understand this from the fundamental theorem of finitely generated abelian groups?

Dietrich Burde
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phy_math
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3 Answers3

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You can see that the (finitely generated) Abelian group $(\mathbb Z\times\mathbb Z)/\langle(n,n)\rangle$ is indeed a product of cyclic groups: $\mathbb Z\times\mathbb Z/n\mathbb Z$, just as the fundamental theorem of finitely generated Abelian groups predicts.

However, the fundamental theorem is wholly unsuitable to predict the classification of this group (or any other finitely generated Abelian group*). It can tell you what it might be, up to isomorphism, but to tell exactly what it is, you need to use different tools, e.g. the ones you actually used to derive your result.

*Except for a cyclic group of prime order.

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I don't see what you mean. You have an automorphism of $\Bbb{Z}^2$ $$(a,b)\to (a-b,b)$$ which sends $(n,n)$ to $(0,n)$ and $\Bbb{Z}^2/\langle (0,n)\rangle$ is not too hard to understand.

This generalizes with the Smith normal form.

reuns
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You can see this product as a direct sum. You can take, by example, the natural isomorphism $$\mathbb{Z} \oplus \mathbb{Z}_n \rightarrow \mathbb{Z}_n \times \mathbb{Z}, x + y \mapsto (x,y)$$

In fact, this is the case whenever you had a finite product! That's because the direct sum is constructed to be a subgroup of the product $$\oplus G_i \subset \Pi G_i $$ such that the elements of the direct sum have just finite non-null coordinates. Well, if you have just a finite number of groups, they are equal!

gabriel fazoli
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