Does $G=HK_1=HK_2$ imply $K_1=K_2$ in a locally cyclic group? Let $G$ be a locally cyclic group and $H,K_1,K_2\le G$ be subgroups such that $G=HK_1=HK_2$ and $H\cap K_1=H\cap K_2=\{1\}$.
Is $K_1=K_2$?
 A: Let $G'$ be a finitely generated subgroup of $G$. Let $H'=H\cap G'$, $K_1' = G' \cap K_1$, and $K_2' = G' \cap K_2$. In a locally cyclic group the lattice of subgroups is distributive, hence
$$ G' = G' \cap HK_1 = (G' \cap H)(G'\cap K_1) = H'K_1',$$
and similarly $G'=H'K_2'$.
Now since $H\cap K = \{1\}$, we have $H' \cap K'=\{1\}$ and so $H'K_1' \simeq H' \times K_1'$ and similarly $H'K_2' \simeq H' \times K_2'$.
The groups $H'$, $K_1'$ and $K_2'$ are all cyclic, being subgroups of the cyclic group $G'$. Therefore we have
$$H_1 \simeq C_h, K_1' \simeq C_a, K_2' \simeq C_b,$$
for some integers $h$, $a$, $b$, where $C_n$ denotes the cyclic group of order $n$. Now from $H'K_1' = H'K_2'$ we get
$$ H' \times K_1' \simeq H' \times K_2' \Longrightarrow C_h \times C_a \simeq C_h \times C_b.$$
Comparing orders we obtain $a=b$, so that $K_1'$ is isomorphic to $K_2'$. Furthermore, since $G'\simeq C_h \times C_a$ is itself a cyclic group, necessarily $(h,a)=(h,b)=1$. It follows that any isomorphism $\phi: C_h \times C_a \rightarrow C_h \times C_a$ must restrict to isomorphisms $\phi_1: C_h \times \{1\} \rightarrow C_h \times \{1\}$ and $\phi_2: \{1\} \times C_a \rightarrow \{1\} \times C_b$. In particular, applying this to the identity map $H'K_1' \rightarrow H'K_2'$, we get $K_1' = K_2'$.
Now consider the filtered direct system $\{G_\alpha\}_{\alpha \in I}$ of all finitely generated subgroups of $G$. Then $G = \varinjlim G_\alpha$. For each $\alpha$ we have $G_\alpha \cap K_1 = G_\alpha \cap K_2$, so in the limit $K_1=K_2$.
A: As I was examining the question and the other answer to the question, I remembered a theorem: the last part of theorem 5.1 in this book:
http://books.google.com/books?id=Ll0JXd11SW0C&pg=PA67#v=onepage&q&f=false
"$G=$" and "$=\{1\}$" in the question are unnecessary.
