# How to find invariant factor of this finitely generated group

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..

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

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 $$\mathbb{Z}\oplus\mathbb{Z}$$.

• 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)? .
– MLe
Commented Sep 22, 2023 at 3:45
• Oh, I found! Thanks a lot!
– MLe
Commented Sep 22, 2023 at 3:56
• @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? Commented Sep 22, 2023 at 4:23
• 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.
– MLe
Commented Sep 22, 2023 at 4:37