What are the symmetry elements of the isomers of dichlorocyclobutane?

Since you have not specified which isomers, I will simply list four of them (assuming they are on different carbons):

From left to right, we have the cis-1,3, trans-1,3, trans-1,2, and cis-1,2 isomers. I will label them from left to right, isomers (1), (2), (3), and (4).

(1): #E#, #C_2#, #sigma_v(xz)#, #sigma_v(yz)#
(2): #E#, #C_2#, #i#, #sigma_h#
(3): #E, C_2#
(4): #E, sigma_h#

So basically:

(1): Identity, one rotation axis, two reflection planes
(2): Identity, one rotation axis, one point of inversion, one reflection plane
(3): Identity, one rotation axis
(4): Identity, one reflection plane

DISCLAIMER: Long answer! Possibly difficult to visualize...

The possible, primary symmetry elements overall (for any compound) are:

• #E#, the identity element, for completeness's sake.
• #C_n#, the principal rotation axis, where a rotation of #360^@/n# about this axis returns the original molecule. This is defined to require the largest rotation angle to return the original molecule.
  We define this to be along the #bb(z)# axis.
• #C_n'#, any other rotation axis, defined similarly; just something that isn't the previously-defined #C_n#. It could be perpendicular.
• #sigma_v#, the vertical reflection plane, colinear with the principal rotation axis #C_n#. Generally, crosses through atoms.
• #sigma_h#, the horizontal reflection plane, generally on the plane of a cyclic molecule that is perpendicular to the #C_n# axis.
  If no #C_n# axis exists except for the trivial #C_1# (#C_1 = E#), then we assign a reflection plane as #sigma_h#.
• #sigma_d#, the dihedral reflection plane, generally in between two atoms, bisecting a bond (this is not common unless #sigma_v# and #sigma_h# are also identified).
• #i#, the point of inversion. Basically you take the coordinates #(x,y,z)# and swap them with the coordinates #(-x,-y,-z)#.

Obviously, all four isomers have #bb(E)#, because they are all themselves. That aside...

ISOMER (1)

• There is one #C_n# axis through the ring, through the plane of the screen. That is a #bb(C_2)# principal rotation axis, along the #z#-axis. Therefore, the plane of the ring is the #xy#-plane.
• There is one #bb(sigma_v(xz))# vertical reflection plane bisecting the molecule through carbon-1 and carbon-3 (perpendicular to the ring).
• There is one more #bb(sigma_v(yz))# vertical reflection plane bisecting the molecule through carbon-2 and carbon-4 (perpendicular to the ring).

That, by the way, assigns this molecule to a #bb(C_(2v))# point group.

ISOMER (2)

Compared to (1), there is no longer a #sigma_v(yz)# reflection plane through carbon-2 and carbon-4, and there is no longer a #C_2# axis through the plane of the screen.

• There is one #bb(C_2)# axis bisecting the molecule through carbon-2 and carbon-4. We now define that as its #z#-axis. Let the plane of the molecule be the #yz#-plane, then (so that the #x# axis is through the plane of the screen).
• There is one point of inversion, #bb(i)#. Try taking carbon-1 and carbon-3, and switching their coordinates. Your chlorines would exactly swap places.
• There is one #bb(sigma_h(xz))# vertical reflection plane bisecting the molecule through carbon-1 and carbon-3 (perpendicular to the ring). Since it is perpendicular to #C_2#, it is still considered "horizontal". We've just reoriented our axes.

That, by the way, assigns this molecule to a #bb(C_(2h))# point group.

ISOMER (3)

Compared to (2), there is no point of inversion #i#.

Using the same reasoning as in (1) and (2):

• There is a #bb(C_2)# principal rotation axis bisecting the #C_1-C_2# bond, and we define that as the #z#-axis since #n# is as low as possible (#C_1 = E#).

And I think that's it... I cannot find any reflection planes. There is no point of inversion as I mentioned before, since the chlorines are both on the same side.

So this is, by the way, assigned to a #bb(C_2)# point group, one of low symmetry.

ISOMER (4)

Compared to (3), there is actually a reflection plane, but no #C_n# axis...

• All I see is a #bb(sigma_h)# reflection plane, since there is no #C_n# axis I can find. This is bisecting the molecule through the #C_1-C_2# bond (perpendicular to the plane of the screen).

So this is, by the way, assigned to the #bb(C_s)# point group, which is of low symmetry.

Does this make sense? It may require you to pull out a model kit to visualize this.