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Is the following: $$(\mathbb{Z}\oplus \mathbb{Z}) / \mathbb{Z} = \mathbb{Z}$$

just the informal way of going through the isomorphism of $\mathbb{Z}$ with the group $$\tilde{\mathbb{Z}} := \{ (z,0) | z \in \mathbb{Z} \}$$ i.e., formally:

$$(\mathbb{Z}\oplus \mathbb{Z}) / \tilde{\mathbb{Z}} = \tilde{\mathbb{Z}} \sim \mathbb{Z}$$

Anon
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    It is not true that "$(\mathbb{Z} \oplus \mathbb{Z})/\mathbb{Z} =\mathbb{Z}$," for two reasons: firstly, because it should be isomorphism, rather than equality, and secondly, because even at the level of isomorphism it is false without specifying which subgroup isomorphic to $\mathbb{Z}$ you are talking about. – Stephen Nov 08 '22 at 15:23
  • See for example this post. – Dietrich Burde Nov 08 '22 at 15:32
  • So you mean, for example, that if I'm referring to ${(z,z) | z \in \mathbb{Z}} \sim \mathbb{Z}$ then it won't hold (EDIT: it does as in the comment above this), but if it's ${(z,0) | z \in \mathbb{Z}}$ or ${(0,z) | z \in \mathbb{Z}}$ then it does? – Anon Nov 08 '22 at 15:32
  • For the cyclic subgroup generated by $(1,1)$ it does hold, see the duplicate. You can use this for your case, too. – Dietrich Burde Nov 08 '22 at 15:33
  • Right - just saw, thanks. Any obvious counter example I'm missing? – Anon Nov 08 '22 at 15:34
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    Oh yes, for $(2,2)$ it is false, for example. See this post. You should read these posts a bit. – Dietrich Burde Nov 08 '22 at 15:35
  • Thanks, they did not come up in my search probably due to notation. Will close the question now - sorry for the potential duplicate. – Anon Nov 08 '22 at 15:36

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Suppose $R$ is a commutative ring with identity and $U$ and $V$ are $R-$ modules. Then $U \oplus V$ is an $R-$ module and the map $$\theta :U \rightarrow U \oplus V$$ defined by $$\theta(u)=(u,0)$$ is a monomorphism. Then $(U \oplus V)/\theta(U)$ is isomorphic to $V.$ You would find it informative to read up on split short exact sequences in a beginning book on homological algebra.

P. Lawrence
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