We are told that G is a cyclic group of order 136. How many elements are in the subgroup $\langle g^{30}\rangle$ ? How many subgroups contain this subgroup?
I used the extended euclidean algorithm and found that the gcd(136,30)=2. Then plugged that into the equation $ord(g^k)$ = n/gcd(136,30) and got 136/2= 68. I believe this means there are 68 elements in the subgroup $<g^(30)>$, is that correct? However, I am not sure what the second question is asking. Any help would be appreciated.
I've just had a thought...since gcd=2, would it have something to do with $g^2$...as in would it be the number of groups that are multiples of 2 up to 136 (2,4,6,...,134)
Rounding up your calculation ( and assuming you actually meant $\;G=\langle g\rangle\;$ ), observe that $\;\langle g\rangle=\langle g^{15}\rangle\;$ , since $\;gc d(136,15)=1\;$ , and thus:
$$g^{30}=\left(g^{15}\right)^2\implies ord(g^{30})=ord(g^2)=68$$
and you can save a lot of annoying calculations.
Now, since it is given $\;G\;$ is cyclic, then it has one unique subgroup of each order dividing $\;136\;$ (and, BTW, all of them cyclic as well...), so all you have to do now is to calculate how many divisors $\;136\;$ has...