# two isomorphic groups

The set of $3\operatorname{x}3$ permutation matrices form a group under matrix multiplication. This example demonstrates that fact and develops the multiplication table and compares it to $S_{3}$. Although there are alternative ways to fill in the table, this example serves to help the beginner. Here we will see that the two groups have the same structure. We begin by defining the elements of our group.

$\begin{array}[]{cc}\hskip 128.037402ptI=\left(\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array}\right)\par &\par R=\left(\begin{array}[]{ccc}1&0&0\\ 0&0&1\\ 0&1&0\end{array}\right)\\ \\ \par \hskip 128.037402pt\par A=\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right)\par &\par S=\left(\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\end{array}\right)\\ \\ \par \hskip 128.037402pt\par B=\left(\begin{array}[]{ccc}0&0&1\\ 1&0&0\\ 0&1&0\end{array}\right)\par &\par T=\left(\begin{array}[]{ccc}0&1&0\\ 1&0&0\\ 0&0&1\end{array}\right)\par \end{array}$

Here, our group is just $P_{3}=\{I,A,B,R,S,T\}$. Now, we can start to multiply and then fill in the table. First, we calculate the square of each elements.

 $A^{2}=\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right)\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right)=\left(\begin{array}[]{ccc}0&0&1\\ 1&0&0\\ 0&1&0\end{array}\right)=B$
 $B^{2}=\left(\begin{array}[]{ccc}0&0&1\\ 1&0&0\\ 0&1&0\end{array}\right)\left(\begin{array}[]{ccc}0&0&1\\ 1&0&0\\ 0&1&0\end{array}\right)=\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right)=A$
 $R^{2}=\left(\begin{array}[]{ccc}1&0&0\\ 0&0&1\\ 0&1&0\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\ 0&0&1\\ 0&1&0\end{array}\right)=\left(\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array}\right)=I$
 $S^{2}=\left(\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\end{array}\right)\left(\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\end{array}\right)=\left(\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array}\right)=I$
 $T^{2}=\left(\begin{array}[]{ccc}0&1&0\\ 1&0&0\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}0&1&0\\ 1&0&0\\ 0&0&1\end{array}\right)=\left(\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array}\right)=I$

Now starting with the upper left $3\operatorname{x}3$ block, we go through the table.

 $AB=\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right)\left(\begin{array}[]{ccc}0&0&1\\ 1&0&0\\ 0&1&0\end{array}\right)=\left(\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array}\right)=I$
 $BA=\left(\begin{array}[]{ccc}0&0&1\\ 1&0&0\\ 0&1&0\end{array}\right)\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right)=\left(\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array}\right)=I$

We can complete the upper left $3\operatorname{x}3$ block of the table and complete diagonal using the above values. We note that no row or column can have a repeated elements which follows from the of a group. Next, we work on the upper right $3\operatorname{x}3$ block of the table.

 $AR=\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\ 0&0&1\\ 0&1&0\end{array}\right)=\left(\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\end{array}\right)=S$
 $AS=\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right)\left(\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\end{array}\right)=\left(\begin{array}[]{ccc}0&1&0\\ 1&0&0\\ 0&0&1\end{array}\right)=T$

Now we can complete the upper right 3 x 3 block of the table. Next, we work on the lower left $3\operatorname{x}3$ block of the table.

 $RA=\left(\begin{array}[]{ccc}1&0&0\\ 0&0&1\\ 0&1&0\end{array}\right)\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right)=\left(\begin{array}[]{ccc}0&1&0\\ 1&0&0\\ 0&0&1\end{array}\right)=T$
 $RB=\left(\begin{array}[]{ccc}1&0&0\\ 0&0&1\\ 0&1&0\end{array}\right)\left(\begin{array}[]{ccc}0&0&1\\ 1&0&0\\ 0&1&0\end{array}\right)=\left(\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\end{array}\right)=S$
 $SA=\left(\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\end{array}\right)\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right)=\left(\begin{array}[]{ccc}1&0&0\\ 0&0&1\\ 0&1&0\end{array}\right)=R$

Now we can complete the lower left $3\operatorname{x}3$ block of the table. Finally, we work on the lower right $3\operatorname{x}3$ block of the table.

 $RS=\left(\begin{array}[]{ccc}1&0&0\\ 0&0&1\\ 0&1&0\end{array}\right)\left(\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\end{array}\right)=\left(\begin{array}[]{ccc}0&0&1\\ 1&0&0\\ 0&1&0\end{array}\right)=B$
 $SR=\left(\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\ 0&0&1\\ 0&1&0\end{array}\right)=\left(\begin{array}[]{ccc}0&1&0\\ 1&0&0\\ 0&0&1\end{array}\right)=A$

This completes the multiplication and the table is given below.

$\begin{array}[]{|c||c|c|c|c|c|c|}\hline&I&A&B&R&S&T\\ \hline\hline I&I&A&B&R&S&T\\ \hline A&A&B&I&S&T&R\\ \hline B&B&I&A&T&R&S\\ \hline R&R&T&S&I&B&A\\ \hline S&S&R&T&A&I&B\\ \hline T&T&S&R&B&A&I\\ \hline\end{array}$

Next, we want to compare this table to the symmetric group $S_{3}$. We begin as before by defining the elements as follows.

 $e={1\ 2\ 3\choose 1\ 2\ 3}\hskip 56.905512ptr={1\ 2\ 3\choose 2\ 1\ 3}$
 $a={1\ 2\ 3\choose 2\ 3\ 1}\hskip 56.905512pts={1\ 2\ 3\choose 3\ 2\ 1}$
 $b={1\ 2\ 3\choose 3\ 1\ 2}\hskip 56.905512ptt={1\ 2\ 3\choose 1\ 3\ 2}$

The multiplication table for this group is obtained within the entry symmetric group on three letters. The table is:

$\begin{array}[]{|c||c|c|c|c|c|c|}\hline&e&a&b&r&s&t\\ \hline e&e&a&b&r&s&t\\ \hline a&a&b&e&s&t&r\\ \hline b&b&e&a&t&r&s\\ \hline r&r&t&s&e&b&a\\ \hline s&s&r&t&a&e&b\\ \hline t&t&s&r&b&a&e\\ \hline\end{array}$

Define the following homomorphism $\varphi\colon S_{3}\to P_{3}$ by the following:

$\varphi(e)=I$;

$\varphi(a)=A$;

$\varphi(b)=B$;

$\varphi(r)=R$;

$\varphi(s)=S$;

$\varphi(t)=T$.

Since $\varphi$ is a bijection, we conclude that $P_{3}$ and $S_{3}$ are isomorphic.

Title two isomorphic groups TwoIsomorphicGroups 2013-03-22 15:52:28 2013-03-22 15:52:28 Wkbj79 (1863) Wkbj79 (1863) 10 Wkbj79 (1863) Example msc 20B30