Given: the abelian group Z*13 and the element, 5, of Z*13. Find the index t of the cyclic subgroup <5>. List the t distinct elements of <5> and use them to fill in a t-by-t multiplication table for G/<a> (G mod <a>).

Then find either an integer l such that G/<a> is isomorphic to Z_{l}or integers m and n such that G/<a> is isomorphic to Z_{m}X Z_{n}. Justify your answer.

Isn't Z*13={1,2,3,4,5,6,7,8,9,10,11,12} ?

Then isn't <5> = {5, 12, 8, 1}?

Since Z*13 has 12 elements wouldn't the index t of <5> be 3 since <5> has 4 elements?

I'm confused because <5> doesn't generate Z*13, so it isn't cyclic? Unless I'm making a mistake?? Then how would I do the multiplication table?

--I just did some more work and I believe now that <5> is a cyclic subgroup. I believe my error was taking into consideration that it is a subgroup, and cannot generate all of Z*13 necessarily. I then found that <1> = {1}, <12> = {12, 1} and <8> = {8, 12, 5, 1}. So would the index t be 2 then or what? Still confused with the multiplication table and the "t distinct elements" of <5>

Finally, any help with the isomorphism? Please and thank you!