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.