Finite Group Z2Z2

One of the two groups of Order 4. The name of this group derives from the fact that it is a Direct Product of two $Z_2$ Subgroups. Like the group $Z_4$, $Z_2\otimes Z_2$ is an Abelian Group. Unlike $Z_4$, however, it is not Cyclic. In addition to satisfying ${A_i}^4=1$ for each element $A_i$, it also satisfies ${A_i}^2=1$, where 1 is the Identity Element. Examples of the $Z_2\otimes Z_2$ group include the Viergruppe, Point Groups $D_2$, $C_{2h}$, and $C_{2v}$, and the Modulo Multiplication Groups $M_8$ and $M_{12}$. That $M_8$, the Residue Classes prime to 8 given by $\{1, 3, 5, 7\}$, are a group of type $Z_2\otimes Z_2$ can be shown by verifying that

$$1^2=1 \quad 3^2=9\equiv 1 \quad 5^2=25\equiv 1 \quad 7^2=49\equiv 1\ ({\rm mod}\ 8)$$ (1)

and

$$3\cdot 5=15\equiv 7 \quad 3\cdot 7=21\equiv 5 \quad 5\cdot 7=35\equiv 3\ ({\rm mod}\ 8).$$ (2)

$Z_2\otimes Z_2$ is therefore a Modulo Multiplication Group.

The Cycle Graph is shown above, and the multiplication table for the $Z_2\otimes Z_2$ group is given below.

$Z_2\otimes Z_2$	1	$A$	$B$	$C$
1	1	$A$	$B$	$C$
$A$	$A$	1	$C$	$B$
$B$	$B$	$C$	1	$A$
$C$	$C$	$B$	$A$	1

The Conjugacy Classes are $\{1\}$, $\{A\}$,

$$A^{-1}AA = A$$ (3)

$$B^{-1}AB = A$$ (4)

$$C^{-1}AC = A,$$ (5)

$\{B\}$,

$$A^{-1}BA = B$$ (6)

$$C^{-1}BC = B,$$ (7)

and $\{C\}$.

Now explicitly consider the elements of the $C_{2v}$ Point Group.

$C_{2v}$	$E$	$C_2$	$\sigma_v$	$\sigma_v$
$E$	$E$	$C_2$	$\sigma_v$	$\sigma_v'$
$C_2$	$C_2$	$E$	$\sigma_v'$	$\sigma_v$
$\sigma_v$	$\sigma_v$	$\sigma_v'$	$E$	$C_2$
$\sigma_v'$	$\sigma_v'$	$\sigma_v$	$C_2$	$E$

In terms of the Viergruppe elements

$V$	$I$	$V_1$	$V_2$	$V_3$
$I$	$V_1$	$V_2$	$V_3$	$V_4$
$V_1$	$V_1$	$I$	$V_3$	$V_2$
$V_2$	$V_2$	$V_3$	$I$	$V_1$
$V_3$	$V_3$	$V_2$	$V_1$	$I$

A reducible representation using 2-D Real Matrices is

$$1 = \left[\begin{array}{cc}1 & 0 \\ 0 & 1 \end{array}\right]$$ (8)

$$A = \left[\begin{array}{cc}-1 & 0 \\ 0 & -1 \end{array}\right]$$ (9)

$$B = \left[\begin{array}{cc}0 & 1 \\ 1 & 0 \end{array}\right]$$ (10)

$$C = \left[\begin{array}{cc}0 & -1 \\ -1 & 0 \end{array}\right].$$ (11)

Another reducible representation using 3-D Real Matrices can be obtained from the symmetry elements of the $D_2$ group (1, $C_2(z)$, $C_2(y)$, and $C_2(x)$) or $C_{2v}$ group (1, $C_2$, $\sigma_v$, and $\sigma_v'$). Place the $C_2$ axis along the $z$-axis, $\sigma_v$ in the $x$-$y$ plane, and $\sigma_v'$ in the $y$-$z$ plane.

$$1 = E = E = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$ (12)

$$A = R_x(\pi) = \sigma_v = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$ (13)

$$C = R_z(\pi) = C_2 = \left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{array}\right]$$ (14)

$$B = R_y(\pi) = \sigma_v' = \left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right].$$ (15)

In order to find the irreducible representations, note that the traces are given by $\chi(1) = 3, \chi(C_2) = -1,$ and $\chi(\sigma_v) = \chi(\sigma_v') = 1.$ 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 rule 1, we are looking for Positive Integers which satisfy

$${l_1}^2+{l_2}^2+ {l_3}^2+{l_4}^2 = 4.$$ (16)

The only combination which will work is

$$l_1 = l_2 = l_3 = l_4 = 1,$$ (17)

so there are four one-dimensional representations. Rule 2 requires that the sum of the squares equal the Order $h = 4$, so each 1-D representation must have Character $\pm 1$. 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 (rule 3) to build up the other representations. The simplest solution is then given by

$C_{2v}$	1	$C_2$	$\sigma_v$	$\sigma_v'$
$\Gamma_1$	1	1	1	1
$\Gamma_2$	1	$-1$	$-1$	1
$\Gamma_3$	1	$-1$	1	$-1$
$\Gamma_4$	1	1	$-1$	$-1$

These can be put into a more familiar form by switching $\Gamma_1$ and $\Gamma_3$, giving the Character Table

$C_{2v}$	1	$C_2$	$\sigma_v$	$\sigma_v'$
$\Gamma_3$	1	$-1$	1	$-1$
$\Gamma_2$	1	$-1$	$-1$	1
$\Gamma_1$	1	1	1	1
$\Gamma_4$	1	1	$-1$	$-1$

The matrices corresponding to this representation are now

$$1 = \left[\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right]$$ (18)

$$C_2 = \left[\begin{array}{cccc}-1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ (19)

$$\sigma_v = \left[\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right]$$ (20)

$$\sigma_v' = \left[\begin{array}{cccc}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right],$$ (21)

which consist of the previous representation with an additional component. These matrices are now orthogonal, and the order equals the matrix dimension. As before, $\chi(\sigma_v) = \chi(\sigma_v')$.

See also Finite Group Z4