I find on internet this:

$\mathbb{Z}_m \times \mathbb{Z}_n$ is cyclic if and only if $\gcd(m,n)=1$.

Then I do the next steps:

  1. $\gcd(4,12,9)$ is 1. Then I assume that $\mathbb{Z}_4 \times \mathbb{Z}_{12} \times \mathbb{Z}_9$ is cyclic.
  2. I'm trying to find and element $(a,b,c)$, such that $a \in \mathbb{Z}_4,b \in \mathbb{Z}_{12}$ and $c \in \mathbb{Z}_9 $.
  3. $\mathbb{Z}_4$ have two generators: <1>, <3>
  4. $\mathbb{Z}_{12}$ have four generators: <1>, <5>, <7>, <11>
  5. $\mathbb{Z}_{9}$ have four generators:<1>, <2>, <4>, <5>, <7>, <11>

I thought that maybe the generator of $a \in \mathbb{Z}_4,b \in \mathbb{Z}_{12}$ and $c \in \mathbb{Z}_9 $ would be a combination of the other generators. Im traying to find it but I don't get it. Then I suspect that the group is not cyclic.

  • 4
    $\begingroup$ Notice that the theorem you are using only applies to a direct product of two cyclic groups, not three. $\endgroup$ – Morgan Rodgers May 20 '16 at 21:05
  • $\begingroup$ $\Bbb{Z}_9$ appears to be generated by an element which it doesn't even contain! $\endgroup$ – ÍgjøgnumMeg May 20 '16 at 21:11
  • $\begingroup$ This is just a guess, but rather than trying to generalise to gmc$(j,k,l) = 1$, maybe you should've generalised to $g(j,k)$ for every pair $(j,k)$ in your direct product? $\endgroup$ – TastyRomeo May 20 '16 at 21:50
  • $\begingroup$ The concept of two integers being relatively prime can generalize to $n$ integers for $n > 2$ in two ways: being relatively prime as an $n$-tuple, which is what you're saying about the triples $(4,12,9)$ and being pairwise relatively prime (all pairs of two different numbers in the list are relatively prime), which is not true of the triple $(4,12,9)$. Alas, it is being pairwise relatively prime which is what you need for the criterion of a direct product of two cyclic groups being cyclic to extend to more than two cyclic groups. $\endgroup$ – KCd May 20 '16 at 21:57

Clearly $lcm(4,12,9)=12 \cdot 9$ kills every element in $\mathbb{Z}_4 \times \mathbb{Z}_ {12} \times \mathbb{Z}_9$. Therefore, $\mathbb{Z}_4 \times \mathbb{Z}_ {12} \times \mathbb{Z}_9$ is not cyclic because no element has order $4 \cdot 12 \cdot 9$.


In a cyclic group there is at most a subgroup of each order. In your group, on the other hand, there are two subgroups of order $2$.


There are so many easy and direct ways to see this is not true... Here is another one:

Every subgroup of a cyclic group is cyclic. However, by the theorem you mentioned, $\mathbb Z_4 \times \mathbb Z_{12}$ is not cyclic, and it is (trivially isomorphic to) a subgroup of your group.


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