$G$ finite abelian. $G/H$, $G/K$ primary cyclic. $H\cap K=1$. $H\neq 1\neq K$. Is $HK=G$? Given a finite abelian group $G$. $H$ and $K$ subgroups of $G$ such that $G/H$ and $G/K$ are primary cyclic. $H$ and $K$ intersect trivially ($H\cap K=1$),  and $H\neq 1$,  $K\neq 1$.
Is it then true that $HK=G$?
My attempt so far: It's easy to show that $[G:HK] | ([G:H], [G:K])$. So if $G/H$ and $G/K$ are primary cyclic for different primes, then we're done. But I don't know what to do if they're not.
Thank you!
 A: If $a\in G$ is of order $p^n$ for some prime $p$ and $a \notin H$ then $a+H$ is of order $p^d$ for some $d\leq n$ so $G/H$ must be a p-group. 
With this argument we also know that $H$ must contain all elements that are not of order $p^m$ for some m. So if |G| has at least two different prime divisors, then $G/H$ and $G/K$ must be p-group and q-group for different primes (because $H\cap K= 1$). Therefore all the p-sylow groups are contained in at least one of H or K so HK=G.
So we are now remained with the case where $|G|=p^n$. I tried to prove this case (for too much time), but it appears that this is not true. For example, take $G=\mathbb{Z}_2 \times \mathbb{Z}_4$ and let H be the subgroup $\{ (1,0), (0,0) \}$ and $K=\{ (1,2), (0,0) \}$, then $G/H\cong G/K \cong \mathbb{Z}_4 $ and $|HK|=4 < |G|=8$
A: I think this is a counterexample (assuming I haven't messed up something; I spent a fair amount of time trying to prove the result is true, reducing and reducing and reducing, and then the distillation suggested this; I really hope I didn't mess it up):
Take $G=C_4\times C_{16}$, with $C_4$ generated by $x$ and $C_{16}$ generated by $y$. Take $H=\langle (x,y^2)\rangle$ and $K=\langle(x,y^4)\rangle$. Then $G/H$ is generated by the images of $(x,1)$ and $(1,y)$. But $(x,1)H = (1,y^{-2})H=(1,y)^{-2}H$, so $G/H$ is generated by $(1,y)H$ and hence is cyclic (and since $G$ is a $p$-group, necessarily primary cyclic). Similarly, $(x,1)K = (1,y^{-4})K$, so $G/K$ is generated by $(1,y)K$ and hence is cyclic.
Suppose that $(x,y^2)^r = (x,y^4)^s$ for some integers $r$ and $s$; then $r\equiv s\pmod{4}$ and $2r\equiv 4s\pmod{16}$. But if $2r\equiv 4s\pmod{16}$, then $r\equiv 2s\pmod{8}$, hence $s\equiv r\equiv 2s\pmod {4}$.  The only possibility is $s\equiv 0 \pmod{4}$, hence $x^s=1$. But we also now have $4s\equiv 0 \pmod{16}$, so $y^{4s}=1$. Therefore, $(x,y^4)^s = (x^s,y^{4s}) = (1,1)$. Therefore, $H\cap K=\{(1,1)\}$.
However, $H$ and $K$ are both contained in $\langle(x,1),(1,y^2)\rangle$, so $HK$ is a proper subgroup of $G$.
Added: Some comments: your hypothesis imply in any case that $G$ is a product of two primary cyclic groups. To see this, note that if $G/H$ is $p$-primary, and $G/K$ is $q$-primary, then $H$ must contain all $p'$-parts of $G$, and $K$ must contain all $q'$-primary parts of $G$, so $H\cap K=1$ implies that there is no prime which is different to both $p$ and $q$. If $p\neq q$, then the $p$-part of $G$ is a direct sum of $p$-primary cyclic groups, the $q$-part a sum of $q$-primary cyclic groups. Say there are two $p$-primary cyclic groups; then $H$ must contain the entire $q$-part as a subgroup, and have a nontrivial intersection with the $p$-part; since $K$ contains the $p$-part, $H\cap K$ would be nontrivial. Symmetrically with $q$. Thus, if $p\neq q$, then $G=C_{p^a}\times C_{q^b}$, and $H=C_{q^b}$, $K=C_{p^a}$.
If $p=q$, then $G$ is a $p$-group. Then $K \cong K/(H\cap K) = HK/H$, which is a subgroup of $G/H$, so $K$ is cyclic. Symmetrically, $H$ is cyclic. Thus, $G$ can be generated by $2$ elements, so the decomposition of $G$ into a direct factors of cyclic groups has at most two factors. But $G$ cannot be cyclic and and $p$-group, because then any two nontrivial subgroups have nontrivial intersection, contradicting the hypothesis. Thus, $G=C_{p^a}\times C_{p^b}$ in this case as well.
Thus, the conditions require that $G=C_{p^a}\times C_{q^b}$ with $p$ and $q$ primes, possibly equal, and $a$ and $b$ positive.
