Deducing which elements of $\Bbb U_{20}$ generate its subgroups given that we know the subgroup structure of $C_4\times C_2$ Given the direct product $C_2\times C_4$ I know that its subgroups will be of the form $\{(h_i,k_i) | h_i \in C_2,k_i \in C_4\}$ where $h_i,k_i$ form subgroups in $C_2,C_4$ respectively . Consider one of its subgroups say $\{(0,0)(0,2)(1,0)(1,2)\}$ ( Although I mean this question generally I only use this as illustrative example ) . Then given the group of units $\Bbb U_{20}=\{1,3,7,9,11,13,17,19\}$ We know by the Chinese remainder theorem that $\Bbb U_{20}\cong\Bbb U_5\times\Bbb U_4 \cong C_2\times C_4 $. My question is how can we decide which elements of $\Bbb U_{20}$ generate the groups which  correspond to the subgroups of $C_4\times C_2$ by looking at subgroups in the way I first described (i.e of the same form as  $\{(0,0)(0,2)(1,0)(1,2)\}$) ? 
Note: If there's a more efficient way I'm unaware of however I would be glad to hear that also, although I'm not particularly well versed in group theory (just the basics )
 A: Be careful what you mean by "where $h_i, k_i$ form subgroups in $C_2, C_4$ respectively". Let $G$ be the subgroup of $C_2 \times C_4$. If you mean that if you collect all the $k_i$ that appear in any element of $G$, then that set $K$ will be a subgroup of $C_4$, then that's true. But if you mean that $G \cong H \times K$ for subgroups $H \leq C_2, K \leq C_4$, then that's not necessarily true. Consider $G = \{(0,0), (1,1), (0,2), (1,3)\}$.
Anyway, what you're really asking is how to explicitly compute the images of the isomorphism $\mathbb{U}_{20} \cong C_2 \times C_4$ other than explicitly checking for a residue class of multiplicative order 4, and then another of order 2 not a power of the first. 
There isn't going to be a good way to do this. In fact, an easier problem is similarly hard. We know abstractly that for any prime $p$, $\mathbb{U}_p \cong C_{p-1}$, but figuring out which residue classes generate this cyclic group is called finding a primitive element, and there's no better algorithm or pattern known in general other than guessing and checking the order of each element. 
