cyclic subgroups and generators 
g is a generator of the cyclic group C45. Sketch the lattice of
  subgroups of $C_{45} $ For each subgroup, state its order and give a
  generator in terms of g. You do NOT need to list the elements of the
  subgroups.

I know that each generator of the group  has an order that is a divisor of 45, 
so all divisors in c_{45} are 1 , 3 ,5 ,9 , 15 ,45 
I dont understand how to draw the lattice? 
which is :

Can someone please explain this picture to me? 
Thank you. 
 A: The lattice gives you all subgroups of $C_{45}$ and their pairwise subgroup relationships. A line connecting two groups means the bottom-most group is a subgroup of the top-most one (and there isn't one "in between"). So $C_3$ is subgroup of $C_9$ and $C_{15}$ (and therefore of $C_{45}$), but not of $C_5$. There is a path from group $H$ to group $G$ only going upward if and only if $H$ is a subgroup of $G$.
For a finite cyclic group $G = \langle g \rangle$ of order $n$, we can find all subgroups by looking at the divisors of $n$. For every $d \;\vert\; n$ we have $H = \langle g^{d} \rangle$ is a cyclic subgroup of order $\frac{n}{d}$ (so isomorphic to $C_{n/d}$).
Hence, to draw the lattice yourself, look at the divisors of $n = 45$. Each of them will give you a subgroup. Note that since all subgroups are also cyclic, if we have $k \;\vert\; d \;\vert\; n$ then $\langle g^d \rangle \subset \langle g^k \rangle \subset \langle g \rangle$ is a tower of subgroups. To see which lines to connect to $G$, we want the prime divisors $d$ of $n$. Otherwise, as we just saw there will be another subgroup in between $\langle g^d \rangle$ and $G$. You can then repeat this procedure for the other subgroups, or alternatively for a subgroup $\langle g^d \rangle$ look at the divisors $d' \;\vert\; n$ that add one prime factor to $d$, i.e. $d' = d \cdot p$. All such $\langle g^{d'} \rangle$ will be the "immediate" subgroups (hence neighbours in the lattice) of $\langle g^d \rangle$.
