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Someone asked me about what is really the quotient $m\mathbb Z/n\mathbb Z$ looks like when $m|n$. Unfortunately, I did some handy calculations for him to convince him, but it didn't work. May someone help me to have a solid explanation about this group? Thank you!

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2 Answers 2

up vote 6 down vote accepted

If $m|n$ then $n\mathbb{Z}\lhd m\mathbb{Z}$. To see it's structure, look at the homomorphism $\varphi:m\mathbb{Z}\to \mathbb{Z}/\frac nm\mathbb{Z}$ defined by $\varphi(k)=\overline{\frac1m\cdot k}$ (the coset of $\frac1m\cdot k$). You can check that $\varphi$ is an epimorphism (onto) and $\ker(\varphi)=n\mathbb{Z}$. Hence, by the first isomorphism theorem, $m\mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/\frac nm\mathbb{Z}$

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$nZ$ is a normal subgroup of $mZ$. Thus, $mZ/nZ$ is the factor group.

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Thank you but I wanted an answer in detailed. – Basil R Dec 1 '12 at 9:24

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