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I know if $(m, n)=1$ then $\mathbb Z/m\mathbb Z \times\mathbb Z/n\mathbb Z$ is isomorphic to $\mathbb Z/\operatorname{lcm}(m,n)\mathbb Z$. Is it true for all $m,n$? I want to understand the structure of $\mathbb Z/6\mathbb Z \times \mathbb Z/4\mathbb Z$. How many and order of cyclic subgroup this group has? I tried one cyclic subgroup of order $12$ and generated by $(1,1)$. Am I right? What about other cyclic subgroups?

Michael Albanese
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Arun Sharma
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    It is not. Look at the order of the sets. One is $mn$ and the other is $lcm(m,n)$ and these two quantities are not same, in general. – joy May 13 '16 at 17:48
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    For example, take $m=2=n$. – lulu May 13 '16 at 17:54
  • @joy Ok...I have understood. So, order of $\mathbb Z/m \mathbb Z \times \mathbb Z/n\mathbb Z$ is always mn ? – Arun Sharma May 13 '16 at 18:00
  • It's probably helpful to note $\mathbb{Z}_{n} \cong \mathbb{Z}/n\mathbb{Z}$. You're probably aware, but I find it a simpler notation.You probably don't have the Fundamental Theorem of Finite Abelian Groups at your disposal, but it looks like you'll be there soon. http://demonstrations.wolfram.com/TheFundamentalTheoremOfFiniteAbelianGroups/ – ShawSa May 13 '16 at 18:38
  • @ShawSa The problem with that notation, especially when $n$ happens to be prime, is that $\Bbb Z_p$ also denote the $p$-adic integers. – Arthur May 13 '16 at 19:54
  • Anyway, ${\mathbb Z}_n = {\mathbb Z}/n{\mathbb Z}$ is a definition not an isomorphism. – Derek Holt May 13 '16 at 20:01
  • @DerekHolt I'll take your word for it that some can choose it as a definition. My textbook (Contemporary Abstract Algebra by Joseph Gallian) defines $\mathbb{Z}_n$ with integer elements (opposed to equivalence classes), and then later asserts that it is isomorphic to $\mathbb{Z}/n\mathbb{Z}$. I suppose it doesn't make much difference either way, unless you wanted to look at sets like $\mathbb{Z}\cap\mathbb{Z}_n$, but I don't think this is very common. – ShawSa May 14 '16 at 03:44

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As already pointed out in the comments, the statement is false.

If $m$ and $n$ are coprime, then $\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/mn\mathbb{Z}$. Using this fact, we see that

$$\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z} \cong (\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z})\times\mathbb{Z}/4\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}\times(\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/12\mathbb{Z}.$$

In general, $\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z} \cong\mathbb{Z}/\operatorname{lcm}(m,n)\mathbb{Z}\times\mathbb{Z}/\operatorname{gcd}(m,n)\mathbb{Z}$.

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Michael Albanese
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  • So, the order of group is always $mn$? – Arun Sharma May 13 '16 at 18:09
  • Yes. The elements of the group $\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$ are of the form $(a, b)$ where $a \in \mathbb{Z}/m\mathbb{Z}$ and $b \in \mathbb{Z}/n\mathbb{Z}$. As there are $m$ choices for $a$ and $n$ choices for $b$, there are $mn$ possibilities for $(a, b)$. – Michael Albanese May 13 '16 at 18:20
  • yes true. What about cyclic subgroups in the example ? – Arun Sharma May 13 '16 at 18:22