# Is $\mathbb{Z}_4 \times \mathbb{Z}_ {12} \times \mathbb{Z}_9$ cyclic?

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.

• Notice that the theorem you are using only applies to a direct product of two cyclic groups, not three. – Morgan Rodgers May 20 '16 at 21:05
• $\Bbb{Z}_9$ appears to be generated by an element which it doesn't even contain! – ÍgjøgnumMeg May 20 '16 at 21:11
• 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? – TastyRomeo May 20 '16 at 21:50
• 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. – 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 $2$.
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.