Understanding how to find the cosets of a group I'm trying to understand how cosets work. I came across with the following question:

Check if there is a subgroup of order $2$ of $U_{20}$ (Euler group) and find its cosets.

In the solution they found that $U_{20}$ has three subgroups of order $2$: $\langle 9\rangle$, $\langle 11\rangle$ and $\langle 19\rangle$. Later they took the group $\langle 9\rangle$ and try to find its cosets in $U_{20}$ - they just said that the cosets are: $\langle 9\rangle, \{3,7\},\{11,19\},\{13,17\}$ without  explaining why. I'm familiar with the theorem that the left cosets are $gH=\{gh\,:\,h\in H\}$ and right cosets are $Hg=\{hg\,:\,h\in H\}$, but I don't understand how this gives me the solution. I also know that from Lagrange we get:
$$ [U_{20}\,:\,\langle 9\rangle]=\frac{|U_{20}|}{|\langle 9\rangle|}=\frac{8}{2}=4$$
My question is how they understood that the cosets are  $\langle 9\rangle, \{3,7\},\{11,19\},\{13,17\}$?
 A: That is not a theorem but the definition of a coset. They probably did not explain it because they just computed the cosets. For example for $g = 3$ we get $$g \cdot \langle 9 \rangle = 3 \cdot \langle 9 \rangle = 3 \cdot \lbrace 9, 1 \rbrace = \lbrace 3 \cdot 9, 3 \cdot 1 \rbrace = \lbrace 7,3 \rbrace.$$ Now just do the computation for the other elements (not being in the cosets so far) and you will see the four cosets. 
A: Note that $U_{20} = \{1,3,7,9,11,13,17,19\}$ and define $H = \langle 9\rangle = \{1,9\}$. Now, as definitions of cosets you gave, we have (note that all calculations are done in modulo $20$)
$$1H = H,\ 3H = \{3\cdot1,3\cdot9\} = \{3,7\},\ 11H = \{11\cdot1,11\cdot9\} = \{11,19\},\ 13H =  \{13,17\}$$
Note that for instance once we find $3H$ to be $\{3,7\}$, we don't check $7H$ again because they must be the same (since $7 \in 3H$). So these four checks are enough.
A: One coset will be the subgroup itself.
Now take an element of the group that is not in any coset you have so far, for example $3$. Multiply this element with the elements in the subgroup (your group is abelian so you need not worry about left and right cosets here) you will get $\{3,7\}$. Repeat this process, say take $11$. Keep doing this until you have no elements left. 
If you need more guidance let me know.
