Suppose $G$ is an infinite $p$-group and $H$ is a minimal infinite subgroup of $G$ (where $p$ is a given prime). If $H=pH$ then $H$ is divisible.
How can this be? An element of $H$ will only be divisible by a power of $p$, am I right?
EDIT: I've been able to prove any $y\in H$ is divisible by $n$ if $(n,p)=1$ or $n=p^r$ but I fail if $n=tp^s$ with $(t,p)=1$.
How silly I am! Just before reading Derek's last comment I found a complete proof. Let $y\in H$ and $n=tp^s$, with $(t,p)=1$. By the O.P. there is $x\in H$ such that $y=tx$. By the O.P. again there is $z\in H$ such that $x=p^s z$. So $y=t(p^s z)=(tp^s)z=nz$. I can't believe such an innocent statement as $H$ $p$-group and $H=pH$ gives $H$ divisible.
EDIT: Ahhh, in the hypothesis (see O.P.) all that matters is that $G$ is a $p$-group. The other assumptions can be dropped.
By the way, if $G$ is any group, $y\in G$ and $m,n$ divide $y$ then $y$ is divisible by $mn$ as suggested above.