Number of homomorphisms from direct products of $\mathbb{Z}_n$ to $\mathbb{Z}_{18}$

Question

How many homomorphisms are there from $\mathbb Z_3\times \mathbb Z_4\times\mathbb Z_9$ to $\mathbb Z_{18}$. I tried to find possible kernals. The answer is $54$ but I'm getting something else. Can anyone show me some easy way to compute these homomorphisms.

Answer 1

The images of the elements $(1,0,0),(0,1,0)$ and $(0,0,1)$ will determine the homomorphism. Also, those images need orders dividing $3,4,9$ respectively.

There are three choices for the image of $(1,0,0)$, mapping to elements of order $1$, that is, to $e$, or two one of the two elements of order $3$.

Next, there are two choices, since the order of $h(0,1,0)$ has to divide $4$, hence be $1$ or $2$ (there are no elements of order $4$).

Finally, there are nine choices for $h(0,0,1)$. Because the order must divide $9$, hence be $1,3$ or $9$. There are $\varphi(3)=2$ elements of order three, and $\varphi(9)=6$ elements of order nine in $\Bbb Z_{18}$.

Thus we have $2\cdot3\cdot9=54$.