Symmetry element in the coversion of cyclopent-2-enyl anion to pentadienyl anion?
Let's look at the back reaction, the closure of the pentadienyl anion to the cyclopentenyl anion.  Once you understand this you'll see that the exact same reasoning applies to the reverse reaction that you asked about.  So here's a picture of the reaction we're talking about.



And here is a picture of the 5 molecular orbitals for the pentadienyl system



Since the pentadienyl anion has 6 pi electrons, the first 3 molecular orbitals (from the bottom up) are filled, each with two electrons.  That makes the third molecular orbital (the one in the middle of the diagram), let me call it $\ce{\psi}_3$, the highest occupied molecular orbital (HOMO).  The HOMO is so important because it is the orbital that will control our process.  If we look at the symmetry of the $\ce{P}$ orbitals on $\ce{C_1}$ and $\ce{C_5}$ in $\ce{\psi}_3$, we see that in order to get constructive overlap (overlapping the parts of the $\ce{P}$ orbitals with the same sign) as we rotate the $\ce{P}$ orbitals on these two carbons to form the bond that forms the cyclopentenyl ring, we must rotate the tops (or bottoms) of these two $\ce{P}$ orbitals towards each other.  If we were to rotate the top of one and the bottom of the other towards each other we would overlap a plus side of one $\ce{P}$ orbital with the minus side of the other $\ce{P}$ orbital and this would not result in constructive or bonding overlap.  The type of rotation that leads to bonding overlap in this example is termed "disrotatory".  It preserves a plane of symmetry bisecting the pentadienyl system (e.g. it contains $\ce{C_3}$) and is perpendicular to the plane of the screen.  This is the symmetry element that is preserved during the ring closure.  The exact same arguments also explain the ring opening of the cyclopentenyl anion to the pentadienyl anion.