Find [ Z_40 : <[12], [20]> ]?
I know it has something to do with the gcd of 12 and 20 which is 4 but I do not know how to proceed with it and simply it with proper steps.
Thanks for the help!
So, it is "well known" that any subgroup of a cyclic group is cyclic. As you have pointed out, gcd(12, 20)=4. This means that the subgroup generated by 12 and 20 is equal to the subgroup generated by 4, $\displaystyle \langle 12, 20\rangle=\langle 4\rangle$ (this should be in your notes, and is the crux of the argument - if you do not know it, learn it!). So, what is the index of $\displaystyle \langle 4\rangle$ in $\displaystyle \mathbb{Z}_4$? Divide 40 by this order to get your answer (why?).