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This question already has an answer here:

Is quotient group $(\mathbb{Z}\times\mathbb{Z})/(\langle (4,6)\rangle)$ cyclic? I'm not sure but I tried to prove that this is equivalent to $\mathbb{Z}$. It does not work and I'm really stuck now.

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marked as duplicate by Dietrich Burde abstract-algebra Oct 13 '17 at 13:43

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Let $v_1=(2,3)$ and $v_2=(1,1)$. Then $\mathbb{Z}\times\mathbb{Z} = \mathbb{Z}v_1 \oplus \mathbb{Z}v_2$ because $\begin{vmatrix}2 & 3 \\ 1 & 1 \end{vmatrix} = -1$

and so $$ (\mathbb{Z}\times\mathbb{Z})/(\langle (4,6)\rangle) = (\mathbb{Z}\times\mathbb{Z})/(2\mathbb{Z}v_1) = (\mathbb{Z}v_1 \oplus \mathbb{Z}v_2) / (2\mathbb{Z}v_1 \oplus 0) \cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z} $$ is infinite and not cyclic.

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Hint: $2 \cdot\overline{(2,3)} = 0$ in the quotient. So if it were cyclic, it would need to be finite, in particular it can't be isomorphic to $\Bbb Z$.

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  • $\begingroup$ Sorry how is $2 \cdot (<(2,3)>)$ in the quotient? $\endgroup$ – Daniel Li Oct 13 '17 at 0:35
  • $\begingroup$ Sorry, I forgot to denote the equivalence class, I fixed it. $\endgroup$ – MatheinBoulomenos Oct 13 '17 at 0:38
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    $\begingroup$ Expansion on the hint: $\overline{(2,3)}$ is a torsion element of the quotient group as shown above, and $\overline{(1,0)}$ is a non-torsion element. Is it possible for a cyclic group to have both torsion and non-torsion elements? $\endgroup$ – Daniel Schepler Oct 13 '17 at 0:43
  • $\begingroup$ Is it possible to do this without using this "torsion" concept; we have just covered quotient group and Isomorphism theorems. $\endgroup$ – Daniel Li Oct 13 '17 at 0:50
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    $\begingroup$ @Tancredi I did not claim that $\Bbb Z$ is not cyclic. I was impyling that there is a non-zero torsion element. $\endgroup$ – MatheinBoulomenos Oct 13 '17 at 2:43

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