Klein four-group

The Klein four-group is the abelian group with three nonidentity elements, all of order two.

Write the Klein four-group as $\{1,a,b,ab\}$. Its four one-dimensional irreducible representations independently assign signs to $a,b$, so the displayed characters have $p,q\in \{0,1\}$. The character table isEvery conjugacy class is a singleton because the group is abelian. The four rows exhaust the complex irreducibles by the equality between their number and the number of classes. If $H$ is this table, $H{H}^T=4I$ proves character orthogonality. The diagonal three-dimensional representation by half-turns about the coordinate axes is the direct sum of the three nontrivial characters, with total character $(3,-1,-1,-1)$.

On the nonzero real numbers, sign reversal $a$ and reciprocation $b$ are commuting involutions. Their function composition generates exactly $e(x)=x$, $a(x)=-x$, $b(x)=1/x$ and $ab(x)=-1/x$. The relations $a^2=b^2=e$ and $ab=ba$ identify the group with $C_2\times C_2$, the Klein four-group. Its subgroups are the trivial group, its three order-two cyclic subgroups and itself. Excluding zero from the domain is necessary for all four transformations to be defined and invertible.

Let the Klein four-group have generators $X,Y$ and $Z=XY$, with $X^2=Y^2=I$ and $XY=YX$. For every real $p$, the matricesdefine a group representation. Indeed, $D(Y{)}^2=I_2$, and $D(X)$ commutes with every matrix, so the group relations hold.

Set $S=\left(\begin{array}{cc}1& -p/2\\ 0& 1\end{array}\right)$. A direct change of basis givesConsequently all these representations are isomorphic representations, with character values $(2,-2,0,0)$ on $(I,X,Y,Z)$. Each is a direct sum of the two one-dimensional representations whose values on $X$ are $-1$ and whose values on $Y$ are $+1$ and $-1$. A group representation need not be orthogonal; nonzero $p$ changes the basis rather than the represented group.