I solved this one by using the same process we used in class with Z_48. Find the divisors of 20 to find each subgroup, then figure out how to generate the subgroup based on its structure.
I think $ Z_{20} $ has 6 subgroups, generated by $ \overline{0} $, $ \overline{1} $, $ \overline{2} $, $ \overline{4} $, $ \overline{5} $, and $ \overline{10} $. For the second part of the question, G also has 6 subgroups, this time generated by $ \overline{0}, \overline{a}, \overline{2a}, \overline{4a}, \overline{5a}, $ and $ \overline{10a} $. Does this seem right?
