I am required to find the number of homomorphisms from $\mathbb{Z_{3}}\times\mathbb{Z_{4}}\times\mathbb{Z_{9}}$ to $\mathbb{Z_{18}}$ . Here's my attempt:- Writing the group $\mathbb{Z_{3}}\times\mathbb{Z_{4}}\times\mathbb{Z_{9}}$ as $\mathbb{Z_{3}}\times\mathbb{Z_{36}}$ . We know that if $\phi$ is a homomorphism then $o(\phi(a))$ must divide $o(a)$ and $18$ where $a\in \mathbb{Z_{3}}\times\mathbb{Z_{36}}$ . If a is of the form $(\bar{0_{3}},\bar{m})$ Then we see that $\bar{m}$ in $\mathbb{Z_{36}}$ must have order $1,2,3,6,9\,or\,18$. So there are $\phi(1)+\phi(2)+\phi(3)+\phi(6)+\phi(9)+\phi(18)=18$ such elements. Then we look at if $a$ is of the form $(\bar{1_{3}},\bar{m})$ or $(\bar{2_{3}},\bar{m})$ both of them must have equal number of elements so we could just count for one case and double it.
We see that $\bar{1_{3}}$ has order $3$ and so the $\bar{m}$ should be such that lcm of $\bar{1_{3}}$ and $\bar{m}$ is a divisor of 18. But if we see the previous cases of $\bar{m}$ , for each of them the $lcm(\bar{1_{3}},\bar{m})$ always divides $18$.
So by previous argument the number is again $\phi(1)+\phi(2)+\phi(3)+\phi(6)+\phi(9)+\phi(18)=18$.
For $a=(\bar{2_{3}},\bar{m})$ the number would be same as the last case of $(\bar{1_{3}},\bar{m})$.
So by my thinking the answer should be $54$. Is my answer or reasoning at all correct?.Have I missed out some cases or have overcounted anything?. I am a noob here. Sorry if I am offending anyone by posting such questions. I tried to search the website but could not find a similar post. So again I am sorry if this is a duplicate. Any help is appreciated.
HINT: From the universal property of direct products, homomorphisms $\mathbb Z_3\times\mathbb Z_4\times\mathbb Z_9\to\mathbb Z_{18}$ are the same as triples of homomorphisms $(\mathbb Z_3\to\mathbb Z_{18},\mathbb Z_4\to\mathbb Z_{18},\mathbb Z_9\to\mathbb Z_{18})$. The question is thus reduced to this.