Let $H$ be a Hilbert space, $M\subset B(H)$ a von Neumann algebra and $\omega \in H$ a vector. Then $\omega$ is cyclic for $M$ if and only if $\omega$ is separating for $M'$.
I proved "$\implies$". As for the other implication, we're trying to show that $M\omega$ is dense in $H$ given that $\omega$ is separating for $M'$. Let $P$ denote the projection onto the subspace $\overline{M\omega}$. We have that $(I-P)\omega=0$, so if we could show that $(I-P)\in M'$, then separability would imply that $I-P=0$, i.e. $I=P$. In particular, we would have $\overline{M\omega}=H$.
Is it true that $(I-P)\in M'$? I haven't been able to show it.
I am thankful for any help.
Let $X\in M $. Then, for any $Y\in M $ you have $XY\omega\in M\omega $. So $XP=PXP $. If you do this for $X $ selfadjoint and take adjoints on the equality, you get $XP=PX $. As selfadjoints span $M $, you get that $P\in M'$.