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| − | I think your general idea is correct. However, to my understanding, <0> | + | I think your general idea is correct. However, to my understanding, <0> and <a^0> are not a subgroup of either Z or G since there is no divisor of 20 for which 20/k = 0. All subgroups of Z and G must satisfy this condition (Theorem 4.3). <20> and <a^20>, however, would be subgroups of Z and G respectively. Please correct me if I am wrong. |
:--[[Ysuo|Yu Suo]] | :--[[Ysuo|Yu Suo]] | ||
Latest revision as of 19:58, 4 February 2009
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?
I think your general idea is correct. However, to my understanding, <0> and <a^0> are not a subgroup of either Z or G since there is no divisor of 20 for which 20/k = 0. All subgroups of Z and G must satisfy this condition (Theorem 4.3). <20> and <a^20>, however, would be subgroups of Z and G respectively. Please correct me if I am wrong.
- --Yu Suo
