Linear representation theory of generalized dihedral groups

This article discusses the linear representations of the generalized dihedral group corresponding to a finite abelian group $$H$$. This group is defined as:

$$G := \langle H,x \mid x^2 = e, xhx = h{^-1} \ \forall \ h \in H \rangle$$.

The irreducible representations
For the discussion below, we let $$n$$ denote the order of $$H$$. We let $$S$$ denote the set of squares in $$H$$, and $$K = H/S$$ is the quotient group. $$K$$ is an elementary abelian $$2$$-group, and we denote its order by $$2^k$$.

One-dimensional representations
There are $$2^{k+1}$$ of these, described as follows.

Since $$K$$ is an elementary abelian group of order $$2^k$$, it has $$2^k$$ one-dimensional representations. Each of these gives rise to a one-dimensional representation of $$H$$, by composing with the quotient map from $$H$$ to $$K$$. Further, each such representation takes values $$\pm 1$$.

For every such representation $$\rho$$ of $$H$$, there are two corresponding one-dimensional representations of $$G$$ whose restriction to $$H$$ is $$\rho$$: one representation sends the element $$x$$ to $$(1)$$, and the other representation sends $$x$$ to $$-1$$. Thus, we get a total of $$2^{k+1}$$ one-dimensional representations.

Two-dimensional irreducible representations
These two-dimensional representations arise from all the representations of $$H$$ that do not descend to $$K$$. There are $$n - 2^k$$ of these.

Start with any representation $$\rho$$ of $$H$$ that does not have $$S$$ in its kernel. Consider the induced representation to $$G$$. Then, this induced representation is irreducible. Further, the induced representations for two representations are equal if and only if they are complex conjugates of each other -- this can readily be verified by looking at character values.

Since $$\rho$$ does not descend to $$K$$, it is not equal to its complex conjugate. Thus, we obtain $$(n - 2^k)/2$$ inequivalent two-dimensional irreducible representations this way.

Orthogonality relations and numerical checks

 * The total number of irreducible representations is $$2^{k+1} + (n - 2^k)/2 = (n + 3 \cdot 2^k)/2$$, which equals the number of conjugacy classes.
 * The sum of squares of irreducible representations is $$2^{k+1}(1)^2 + (n - 2^k)/2 \cdot 2^2 = 2n$$, which is the order of $$G$$. This confirms the fact that sum of squares of degrees of irreducible representations equals order of group.
 * The degrees of all irreducible representations are $$1$$ or $$2$$. This confirms the fact that degree of irreducible representation divides index of abelian normal subgroup.