We’re now going to use induced representations to find degree two representations for the dihedral group of order We already gave these representations in Example 1 (also see the remark after that!) but we didn’t explain how we got them!

**Example**. Let be the generators of where and Let Show that every induced representation from to has degree two and find all of them.

**Solution**. Since is a cyclic group, it has representations and all of them have degree one (see Example 1). Thus, by the theorem in this post, if is a representation of then Now let be a representation of Then where is an -th root of unity. Let be the induced representation of So where in the tensor product is considered as a left -module and the action is defined by i.e. for all and Now, is a set of all left cosets of in Thus if we let and then is a -basis for Note that corresponds to the standard basis of Now, to find I just need to find, for all the matrix of with respect to We know that an element is either in the form or in the form So we will consider two cases:

*Case 1* . In order to find the matrix of with respect to we only need to find We have

Similarly, since we have

So in this case the matrix of with respect to is

*Case 2* . A similar argument as case 1 shows that the matrix in this case is

In case 1 (and case 2) in the above, the reason that we were allowed to move powers of to the right side of tensor product is that the tensor product is over and any power of is in because