Molecular Symmetry and Group Theory 
The multiplication table derived earlier for C_{2v}  The A_{2} irreducible representation  


In the table below, these numbers have been substituted for the operations in row 1 and column 1. Complete the multiplication table using the numbers.
In the point group C_{2v}, σ(xz)C_{2} = σ(yz) which becomes 1 × 1 = 1, so multiplication of the numbers allocated is consistent with the original table.
Is this true for all the numbers in the table? (If not, seek help!)
Now complete the multiplication table using the assignment:
E  C_{2}  σ(xz)  σ(yz) 

1  1  1  1 
In this case, substituting numbers for σ(xz) and C_{2} gives 1 × 1 = 1, but the result of the combined operation σ(xz)C_{2} is σ(yz), which according to the assignment should have the number 1.
The numbers 1 1 1 1 cannot represent the effects of the group operations satisfactorily, and are therefore not an irreducible representation of the C_{2v} point group. There are in fact only four sets of numbers for which the multiplication table works: these are the characters of the four irreducible representations A_{1}, A_{2}, B_{1} and B_{2}.
The number of irreducible representations is always equal to the number of classes of symmetry operations in the group.
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