Why is $U(12) = U_{4} (2)~ U_3(12)$
Attempt: Any subgroup $U_k(n) = \{x \in U(n)~~|~~x \mod k=1 , k ~|~n \}$
Hence : if $U(12) =\{1,5,7,11\}$
then : $U_4(12) =\{1,5\}$ and $U_3(12)=\{1,7\}$
We see that $U_4(12) ~U_3(12)= \{1,5\} \{1,7\} = \{1,5,7,11\} = U(12)$
I have already proved this using the fact that $U_t(st) \approx U(s)$ and that $U_t(st),U_s(st)$ are disjoint from each other when $\gcd (s,t)=1$
But, does there exist a more elementary proof not necessarily using isomorphism . I tried a modular arithematic proof but it's proving to be too tedious and not going anywhere.
Thank you for your help.
"More elementary proof"...than the one you've produced? This is a very basic fact and you've already proved it in a basic way...and modular arithmetic kicks in in the fact that both subgroups are isomorphic to each other and to cyclic group $\;C_2\;$ .
"Another" way, I guess, you could do it: it's easy to check that $\;U(12)\cong V:=C_2\times C_2=$ the Klein group (non-cyclic of order four), and what you've shown above in fact is that the product of $\;U_3(12)\;,\;\;U_4(12)\;$ is (isomorphic to) their direct product as they're clearly normal and their intersection is trivial, thus you've in fact that
$$U(12)=U_3(12)\times U_4(12)=U_3(12)U_4(12)$$
Perhaps you could also be interested in remarking that $\;U_3(12)\,,\,U_4(12)\cong C_2\;$ ...and whatever you use, at the bottom line you will have to use something from group theory as this is a group theoretic question.