# 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.

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

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.

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

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

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

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

$\varphi(e)=I$;

$\varphi(a)=A$;

$\varphi(b)=B$;

$\varphi(r)=R$;

$\varphi(s)=S$;

$\varphi(t)=T$.

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