The finite group is one of the two distinct groups of group order 4. The name of this group derives from the fact that it is a group direct product of two subgroups. Like the group , is an Abelian group. Unlike , however, it is not cyclic.
The abstract group corresponding to is called the vierergruppe. Examples of the group include the point groups , , and , and the modulo multiplication groups and (and no other modulo multiplication groups). That , the residue classes prime to 8 given by , are a group of type can be shown by verifying that
(1)

and
(2)

is therefore a modulo multiplication group.
The cycle graph is shown above. In addition to satisfying for each element , it also satisfies , where 1 is the identity element.
Its multiplication table is illustrated above and enumerated below (Cotton 1990, p. 11).
1  
1  1  
1  
1  
1 
Since the group is Abelian, the conjugacy classes are , , , and . Nontrivial proper subgroups of are , , and .
Now explicitly consider the elements of the point group.
In terms of the vierergruppe elements
has cycle index
(3)

A reducible representation using twodimensional real matrices is
(4)
 
(5)
 
(6)
 
(7)

Another reducible representation using threedimensional real matrices can be obtained from the symmetry elements of the group (1, , , and ) or group (1, , , and ). Place the axis along the zaxis, in the  plane, and in the  plane.
(8)
 
(9)
 
(10)
 
(11)

In order to find the irreducible representations, note that the traces are given by and Therefore, there are at least three distinct conjugacy classes. However, we see from the multiplication table that there are actually four conjugacy classes, so group rule 5 requires that there must be four irreducible representations. By group rule 1, we are looking for positive integers which satisfy
(12)

The only combination which will work is
(13)

so there are four onedimensional representations. Group rule 2 requires that the sum of the squares equal the group order , so each onedimensional representation must have group character . Group rule 6 requires that a totally symmetric representation always exists, so we are free to start off with the first representation having all 1s. We then use orthogonality (group rule 3) to build up the other representations. The simplest solution is then given by
1  
1  1  1  1  
1  1  
1  1  
1  1 
These can be put into a more familiar form by switching and , giving the character table
1  
1  1  
1  1  
1  1  1  1  
1  1 
The matrices corresponding to this representation are now
(14)
 
(15)
 
(16)
 
(17)

which consist of the previous representation with an additional component. These matrices are now orthogonal, and the order equals the matrix dimension. As before, .