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Suppose that $m,n \in \mathbb{Z}$ and $m$ divides $n$. Show that $$\frac{\mathbb{Z}_n}{\mathbb{Z}_m} \cong \mathbb{Z}_\frac{n}{m}$$ I try to use the third isomorphism theorem to show but I don know how to apply it here. Anyone can guide me ?

Idonknow
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5 Answers5

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Hint: First, note that, strictly speaking, $\mathbb Z_m$ is not a subgroup of $\mathbb Z_n$ (unless $n=m$), and what is really meant here is to first identify a certain subgroup of $\mathbb Z_n$ that is isomorphic to $\mathbb Z_m$. If you first make this precise, a very natural choice of a function $f:\mathbb Z_n\to \mathbb Z_{\frac{n}{m}}$ will spring to mind, prompting an application of the (first) isomorphism theorem.

General note: instead of trying to think which of the isomorphism theorems to use, look for the 'evident homomorphism' and apply the (first) isomorphism theorem to it. More often than not, it is easier to find naturally occurring homomorphisms. The second and third isomorphism theorems are immediate consequences of the first one (so in a sense, there is only one isomorphism theorem).

Ittay Weiss
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You can see this problem via another point of view. For any $m$ which divides $n$ we have $$n\mathbb{Z}\lhd m\mathbb{Z}$$ and by letting $$\varphi:m\mathbb{Z}\to \mathbb{Z}/\frac nm\mathbb{Z}$$ with $$\varphi(k)=\big[{\frac1m\cdot k}\big]$$ one could check; $\varphi$ is an epimorphism and so $\ker(\varphi)=n\mathbb{Z}$. And therfore according to first isomorphism theorem, $$m\mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/\frac nm\mathbb{Z}$$ This can lead you to have the conclusion.

Mikasa
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  • Nice perspective! :+) – amWhy Mar 03 '13 at 15:02
  • why $\frac{mZ}{nZ} \cong \frac{Z_n}{Z_m}$? – Idonknow Mar 03 '13 at 16:15
  • @Idonknow Please stop writing $\frac{\mathbb Z_n}{\mathbb Z_m}$! It's quite wrong and anesthetic. (Btw, where have you seen into the answer above $\frac{\mathbb Z_n}{\mathbb Z_m}$? Do you think that $\mathbb Z/\frac{n}{m}\mathbb Z$ is the same as $\frac{\mathbb Z_n}{\mathbb Z_m}$???) –  Mar 03 '13 at 16:32
  • @YACP: I thought $\mathbb{Z}/\frac{n}{m} \mathbb{Z} \cong \mathbb{Z_{\frac{n}{m}}}$? – Idonknow Mar 03 '13 at 16:46
  • @Idonknow Then what have you thought is not the same to what you have wrote! –  Mar 03 '13 at 17:42
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Define $\phi: \mathbb{Z}_n \to \mathbb{Z}_{n}$ such that $$\phi(x)=mx$$ And consider $\ker \phi$ and $\operatorname{Im}\phi$, and apply First isomorphism theorem.

Hanul Jeon
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Hint: A quotient of a cyclic group is cyclic--compare orders.

Alex Youcis
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  • mind to elaborate ? I don understand what you mean – Idonknow Mar 03 '13 at 08:13
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    @Idonknow There is only one cyclic group of a given order $n$--namely, $\mathbb{Z}n$. Now, as I said, the quotient of a cyclic group is cyclic. From this you see that $\mathbb{Z}_m/\mathbb{Z}_n$ is cyclic, and since it is of order $\frac{m}{n}$ the first sentence tells you it must be isomorphic to $\mathbb{Z}{\frac{m}{n}}$. – Alex Youcis Mar 03 '13 at 08:21
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Consider $\mathbb{Z}_{n}$ as the set $\{0, \cdots, n-1\}$ with addition defined modulo $n$. If we have $\mathbb{Z}_{6} = \{0,1,2,3,4,5\}$, we can identify a copy of $\mathbb{Z}_{3}$ inside $\mathbb{Z}_{6}$ as $\{0,2,4\}$. If we mod out by this subgroup, we get the cosets $\{0,2,4\}$ and $\{1,3,5\}$, which we can identify with $\mathbb{Z}_2$.

Let us use this example to motivate the general construction. Given $\mathbb{Z}_n$, the subgroup generated by $\frac{n}{m}$ has order $m$. The quotient subgroup is still cyclic, and is generated by $[1]$, which has order $\frac{n}{m}$, so the quotient is $\mathbb{Z}_{\frac{n}{m}}$.

Elchanan Solomon
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  • how you think of the map ? I mean how do you know we can define the map like this ? $x \rightarrow \frac{x}{m}$? – Idonknow Mar 03 '13 at 08:06