Is every subgroup of the product of two cyclic groups is again a product of two cyclic groups? Well, this is my question.
Is every subgroup of the product of two cyclic groups is again a product of two cyclic groups (maybe one being trivial)?
Thanks!
 A: Yes.
Assume first that $a,b$ in $C_a\oplus C_b$ are prime powers of the same prime $p$.
Then any subgroup $H$ of it is the direct sum of finitely many cyclic $p$-groups. If there are $n$ summands, then there are $p^n-1$ elements of order $p$ in $H$. As there are only $p^2-1$ elements of order $p$ in $G$, we see that there are at most two summands.
Now for the general finite case, we can split cyclic groups of composite orders into their prime power components, see that there are (at most) two summands per prime in $H$, and combine coprime summands again to finally obtain two summands for $H$ itself.

The above works only for the finite case.
To cover all cases, we should especially cover the case $G=\mathbb Z\oplus \mathbb Z$. But once we have that case, we immediately get the result also for the general case: If $G$ is abelean with at most two generators, it is a quotient of $\mathbb Z\oplus\mathbb Z$. A subgroup $H\le G$ maps to a subgroup of $\mathbb Z\oplus\mathbb Z$ under the canonical projection, hence is 8as we will see in a moment) generated by at most two elements, hence $H$ itself is generated by at most two elements (obtained from preimages of the generators in the quotient).
Let $H\le\mathbb Z\oplus\mathbb Z$.
If $H=0$ we are done.
Otherwise let $(a,b)\in H$ be any nonzero element.
Then $f\colon H\to\mathbb Z$, $(x,y)\mapsto ay-bx$ has nontrivial kernel and some $k\mathbb Z$ with $k\ge 0$ as image. If $k=0$, the homomorphism $H\to\mathbb Z$, $(x,y)\mapsto ax+by$ is injective (we have $ay-bx=0$ from $k=0$, hence $ax+by=0$ implies $(a^2+b^2)x=(a^2x+aby)+(b^2x-aby)=0$, hence $x=0$ and similary $y=0$); the $H$ is infinite cyclic or zero.
And if $k>0$, pick $(c,d)\in H$ with $f(c,d)=k$. Then $g\colon H\to\mathbb Z$, $(x,y)\mapsto (ax+by)-\frac{f(x,y)}k(ac+bd)$ is a homomoprhism with $\langle(c,d)\rangle$ as kernel and an infinite cyclic subgrup of $\mathbb Z$ as image, which shows that $H\cong \mathbb Z\oplus\mathbb Z$.
A: No. For example, you can check in GAP or Magma that  SmallGroup(32,9) is a product of cyclic subgroups of order $4$ and $8$, but it has an elementary abelian subgroup of order $8$ which cannot be a product of two cyclic groups.
A: if $x$ is co-prime to its complex n-gon in real space, then yes. A proof is trivial if you examine the Taylor expansion around $x=i$ and then translated through a subtended angle of epsilon.
