# $\mathbb{Z}\oplus \mathbb{Z}\cong \mathbb{Z}\oplus A \implies \mathbb{Z}\cong A$

Let $$A$$ be a $$\mathbb{Z}$$-module and suppose that $$\mathbb{Z}\oplus \mathbb{Z}\cong \mathbb{Z}\oplus A$$. Do we have $$\mathbb{Z}\cong A$$?

I know the result is true if $$A$$ is finitely generated as $$\mathbb{Z}$$-module. Clearly, $$A$$ can be viewed as a submodule of $$\mathbb{Z}\oplus \mathbb{Z}$$.

Maybe a submodule of $$\mathbb{Z}\oplus \mathbb{Z}$$ is finitely generated?

• Just a comment about your last question. The module $\mathbb{Z} \oplus \mathbb{Z}$ is noetherian (it's finitely generated over a noetherian ring). Hence any submodule of $\mathbb{Z} \oplus \mathbb{Z}$ is finitely generated Jun 27, 2020 at 22:05
• In fact, it is more generally true that if $A$ and $B$ are abelian groups and $\mathbb{Z}\oplus A\cong\mathbb{Z}\oplus B$, then $A\cong B$. See, for example, this question, although the case $B=\mathbb{Z}$ that you ask about is easier than the general case. Jun 28, 2020 at 10:09

If $$\Bbb Z \oplus A$$ is finitely generated, then $$A$$ must be finitely generated. So, it indeed holds that if $$\Bbb Z \oplus \Bbb Z \cong \Bbb Z \oplus A$$, then $$A$$ is finitely generated and your previous observation applies. So, $$A \cong \Bbb Z$$.
• To be explicit, if $(n_1,a_1),\cdots,(n_k,a_k)$ generate $\mathbb{Z}\oplus A$, then $a_1,\cdots,a_k$ generate $A$.
• @user745578: Though subgroups of f.g. modules need not be f.g., quotients do; and $A$ is a quotient of $\mathbb{Z}\oplus A$. Jun 27, 2020 at 22:09