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:


*

*$\gcd(4,12,9)$ is 1. Then I assume that $\mathbb{Z}_4 \times \mathbb{Z}_{12} \times \mathbb{Z}_9$ is cyclic.

*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 $.

*$\mathbb{Z}_4$ have two generators: <1>, <3>

*$\mathbb{Z}_{12}$ have four generators: <1>, <5>, <7>, <11>

*$\mathbb{Z}_{9}$ 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. 

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 $2$.
