Finding elements of a particular order in the direct product of abelian groups

Question

Simple question here.

I am looking to prove that the direct product group $$\mathbb{Z}_4 \times\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_5$$ has no element of order $18$. (Where $\mathbb{Z}_n$ represents the group of integers modulo $n$ under addition).

I know that an element in this group of the form $(a,b,c,d)$ will have order $lcm(A,B,C,D)$, where $A$ is the order of $a$ in $\mathbb{Z}_4$, $B$ is the order of $b$ in $\mathbb{Z}_3$, etc.

By Lagrange's theorem, I know that the order of elements in a group must divide the order of the group, so $A$ is $1,2,$ or $4$, $B$ is $1$ or $3$, etc.

Is there a "nice" way to show that there is no $(A,B,C,D)$ will satisfy this without explicitly?

Answer 1

You've said it yourself:

Every element has order dividing $lcm(4,3,3,5)=60$ but $18$ does not divide $60$.