some things you should be aware of (and might try to prove, if you feel ambitious):

1) every subgroup of a cyclic group is cyclic.

2) if G is cyclic of order n, and k is a divisor of n, G has EXACTLY one subgroup of order k.

in particular, (2) indicates your hunch is correct. the subgroups of Z_{30}are:

<0> = {0}

<1> = Z_{30}(and also equals <7>, <11>, <13>, <17>, <19>, <23>, and <29>)

<2> = {0,2,4,6,8,10,12,14,16,18,20,22,24,26,28} (this is also equal to <4>, <8>, <14>, <16>, <22>, 26>, and <28>)

<3> = {0,3,6,9,12,15,18,21,24,27} (this is also equal to <9>, <21>, and <27>)

<5> = {0,5,10,15,20,25} (this is also equal to <25>)

<6> = {0,6,12,18,24} (this is also equal to <12>, <18>, and <24>)

<10> = {0,10,20} (this also equals <20>)

<15> = {0,15}

we can prove some results about D_{2n}, the dihedral group of order 2n. note that this can be presented as <r,s> where:

r^{n}= e

s^{2}= e

sr = r^{-1}s.

you can prove by induction on |k| that:

sr^{k}= r^{-k}s for all integers k.

this means all elements of the form r^{k}s have order 2:

(r^{k}s)^{2}= (r^{k}s)(r^{k}s) = r^{k}(sr^{k})s = r^{k}(r^{-k}s)s = s^{2}= e.

all the other elements are in <r>, and can be analyzed like elements of any cyclic group.