Since the four given symmetry elements are independent and satisfy all the
group conditions, hence they form a group of order 4. Further, there is one horizontal mirror plane perpendicular to 2 fold rotation axis, therefore the point group is
C 2h (2/m).
Example 3 Show that the symmetry elements E, C 2 , r xz , r yz constitute a
group. What is its point group?
Solution: Given: Four symmetry elements are: E, C 2 , r xz , r yz ; point group = ?
Now, let us check that they follow the group conditions.
To check the closure property let us take their products, that is,
C 2 r xz ¼ r yz
C 2 r yz ¼ r xz
r xz r yz ¼ C 2
To check the associative property, let us consider the triple product, that is,
C 2 r xz r yz
À
Á ¼ C 2 r xz
ð
Þr yz
LHS ¼ C 2 r xz r yz
À
Á ¼ C 2 C 2 ¼ E
RHS ¼ C 2 r xz
ð
Þr yz ¼ r yz r yz ¼ E
) LHS ¼ RHS
The given symmetry elements contain one identify elements E, which leaves the
other members unchanged, that is,
E C 2 ¼ C 2 E ¼ C 2
E r xz ¼ r xz E ¼ r xz
E r yz ¼ r yz E ¼ r yz
Every given symmetry elements are found to have its have own inverse, that is,
C 2 C
À1
2 ¼ C 2 C 2 = E
r xz r xz À1 ¼ r xz r xz ¼ E
r yz r yz À1 ¼ r yz r yz ¼ E
Since the four given symmetry elements are independent and satisfy all the
group conditions, therefore they form a group of order 4. Further, there are two
vertical mirror planes and one 2-fold pure rotation axis, therefore the point group is
C 2v (mm2).
6.3 Group (Point) Representation of Symmetry Operations
241