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Is $\mathbb{Z}/m\mathbb{Z}/n (\mathbb{Z}/m\mathbb{Z})=\mathbb{Z}/(m\mathbb{Z} +n \mathbb{Z})$?

Thanks.

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up vote 5 down vote accepted

You are correct. The easy explanation is that you are modding out by $m$, then modding out by $n$, therefore to do it one step, you mod out both of them: i.e. the subgroup generated by both $m$ and $n$.

A calculation based on the isomorphism theorems is to observe $$n(\mathbb{Z} / m\mathbb{Z}) = (n \mathbb{Z} + m \mathbb{Z})/m \mathbb{Z}$$ and so

$$(\mathbb{Z} / m \mathbb{Z}) / n(\mathbb{Z} / m \mathbb{Z}) = (\mathbb{Z} / m \mathbb{Z}) / ((n \mathbb{Z} + m \mathbb{Z})/m \mathbb{Z}) = \mathbb{Z} / (n \mathbb{Z} + m \mathbb{Z}) $$

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Let $\mathbb{Z}/m\mathbb{Z}$ be generated by $a$. Then $n(\mathbb{Z}/m\mathbb{Z})$ is generated by $na$. So the result is $\mathbb{Z}/d\mathbb{Z}$ where $d=\text{gcd}(n,m)$, which is indeed the same as $\mathbb{Z}/(m\mathbb{Z}+n\mathbb{Z})$.

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but $mZ + nZ = dZ$ no? Z/dZ should not be the same as Z/(m/d)Z then... –  fpqc Jan 11 '13 at 7:56
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