Suppose that G is an abelian group of order 66. Show that G is a cyclic group and show that there are proper subgroups $H$ and $K$ of $H$ such that $G\cong G/H \times G/K$
Now, for this question, I would like to use the results of a previous question that I asked; Let $\theta : G \rightarrow G/H \times G/K$ defined by $g\rightarrow (gH)(gK)$, show that $G/(H\cap K) \cong$ to a subgroup of $G/H \times G/K$.
For the first part, we have that $66=2^13^1 11^1$. Therefore, by using Cauchy's theorem, G has elements $x,y,z$ of order $2,3$ and $11$ respectively. These commute and are relatively prime, so the order of $xyz$ is $66$ and G is cyclic. Please let me know if this part follows, I have only seen this being done with two numbers!
Now, let us have the subgroups $K=\langle x, y \rangle$ and $H=\langle z \rangle$. I don't really know where to go from here!