Use the multiplication table you found in Problem 10 to find the classes in the symmetry group of a square. Show that the 2 by 2 matrices you found are an irreducible representation of the group (see Problem 13), and find the character of each class for that representation. Note that it is possible for the character to be the same for two classes, but it is not possible for the character of two elements of the same class to be different.

The square is given.

In abstract algebra, the symmetric group defined over any set is the group whose elements are all the bisections from the set to itself, and whose group operation is the composition of functions.

The square with given have a centre o.

Multiplication table is:

A A B C I F G E D

B B C I A E D GF

DDBEF I I B CA

E E F D G B I A C

F F D G E A C I B

G G E F D C A B I

$0^{\rho }$Rotational matrix is$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$9)°$Rotational matrix islocalid="1659080708028" $A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

$85°$Rotational matrix is$B=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$

$270°$Rotational matrix is $C=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

After reflection through line x -axis, transformation is,

$$\begin{gathered} (x,y)\to (x,-y)=(1-x+0-y,0-x+1-y) \\ =(\begin{array}{c}1\\ 0\end{array} \\ D=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \end{gathered}$$

After reflection through line-axis, transformation is,

$$\begin{gathered} (x,y)\to (-x,y)=(-1·x+0-y,0·x+1-y) \\ =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right) \\ E=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right) \end{gathered}$$

After reflection through diagonal line, transformation is,

$$\begin{gathered} (x,y)\to (y,x)=(0-x+1-y,1-x+0·y \\ F=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \end{gathered}$$

After reflection through diagonal line, transformation is,

$$\begin{gathered} (x,y)\to (-y,-x)=(0·x-1-y,-1·x+0-y)=\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right) \\ G=\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right) \end{gathered}$$

And the multiplication table is:

According to definition of conjugate elements,

The element A is conjugate to B if there exist C in group G such that

$C^{-1}AC=B$

Consider following multiplications to find conjugate elements to A,

$\begin{array}{l}{l}^{-1}Al=A\\ A^{-1}AA=CA\\ A=CB=A\\ B^{-1}AB=BA\end{array}$

And other values are

$$\begin{gathered} B=B \\ C=A \\ {C}^{-1}AC=AA \\ C=A \end{gathered}$$

Then

$$\begin{gathered} l=A \\ {D}^{-1}AD=DA \\ D=DF=C \\ {E}^{-1}AE=EA \\ E=EG=C \\ {F}^{-1}AF=F^{-1} \\ E=F \\ E=C{G}^{-1} \\ AG=GA \\ G=G \\ D=C \end{gathered}$$

So, the set of all conjugate to A is {A<C}

Find the conjugate of all the elements then

Now, the conjugate classes of this group G are {1}{A,C}{B}{D,E}{F,G}

Let group G is reducible. Then, matrix C exist such that

$\begin{array}{l}{C}^{-1}{X}_{i}C=D_i,X_i\in G,\\ D_i{C}^{-1}DC=D_1{C}^{-1}FC\\ =D\end{array}$

As,$D_1,D_2$are diagonal matrices then these are commutates

Hence, assumption that group G is reducible is wrong.

So, the group G is irreducible.

Hence, the conjugate classes of the group G are {1}{A,C}{B}{D,E}{F,G}

The character of class is trace of each matrix of class.

The character group G are 2,0, - 2,0,0.