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

Question

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.

Answer 1

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 $4 \cdot 12 \cdot 9$.

Answer 2

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 $2$.

Answer 3

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.