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I'm trying to solve this problem.

Let $G$ be the set $\Bbb Z^2$ with binary operation defined by $(x_1,y_1)*(x_2,y_2) = (x_1+x_2, y_1+y_2+x_1x_2)$

Show that $(G,*)$ is a finitely generated abelian group and find its invariant factor.

I showed that G is abelian group and finitely generated. (0,0) is the identity and $(x,y)^{-1}=(-x,x^2-y)$ since $(x,y)*(-x,x^2-y)=(x-x,y+x^2-y-x^2)=(0,0)$

Also I found that $\langle (1,0),(0,1)\rangle =G$

Then how can I find invariant factor?

Since this is an infinite group, I can know that a $Z$ factor is contained..

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  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Commented Sep 21, 2023 at 23:33
  • $\begingroup$ @Shaun is a problem in my writing? $\endgroup$
    – MLe
    Commented Sep 22, 2023 at 0:22
  • $\begingroup$ E.g., $\Bbb Z$ for $\Bbb Z$ and $a^{-1}$ for $a^{-1}$. $\endgroup$
    – Shaun
    Commented Sep 22, 2023 at 0:25
  • $\begingroup$ Use $\langle X\rangle$ for $\langle X\rangle$. $\endgroup$
    – Shaun
    Commented Sep 22, 2023 at 0:26
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    $\begingroup$ Second was typo. I didn't know the last one. Thanks. $\endgroup$
    – MLe
    Commented Sep 22, 2023 at 0:35

1 Answer 1

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Any element with nonzero first coordinate has infinite order, since $(a,y)^n$ has first coordinate equal to $na$.

An element with first coordinate zero has finite order only if it is the zero element: $(0,y)^m = (0,my)$.

Therefore, this is a torsionfree abelian group, generated by two elements. It is either cyclic, or isomorphic to $\mathbb{Z}\oplus\mathbb{Z}$.

Now note that no nontrivial power of $(1,0)$ can equal a nontrivial power of $(0,1)$: for the former have nonzero first coordinate, and the latter have first coordinate equal to zero. So the group cannot be cyclic, because in a cyclic group, if $x$ and $y$ are two nontrivial elements, then there exist nonzero integers $m$ and $n$ such that $x^m=y^n$.

Thus, the group is isomorphic to $\mathbb{Z}\oplus\mathbb{Z}$.

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  • $\begingroup$ Thanks for answering. But I didn't understand that a torsionfree abelian group genereated by two elements is isomorphic to a cyclic group or Z⊕Z. Do you use a theorem (which seems what I don't know)? . $\endgroup$
    – MLe
    Commented Sep 22, 2023 at 3:45
  • $\begingroup$ Oh, I found! Thanks a lot! $\endgroup$
    – MLe
    Commented Sep 22, 2023 at 3:56
  • $\begingroup$ @MLe It would appear you have never accepted an answer to any of your questions, even those where you explicitly thank an answerer. Is there a reason why you have not accepted any answer? $\endgroup$ Commented Sep 22, 2023 at 4:23
  • $\begingroup$ I didn't know that such a function exists. I think I have just done but if it hasn't done yet, let me know. $\endgroup$
    – MLe
    Commented Sep 22, 2023 at 4:37

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