Find the permutation representation of $S_3$. I don't know how to write a solution. Is my solution ok? (Herstein "Topics in Algebra 2nd Edition") I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
The following problem is Problem 20 on p.81 in this book:

Let $G$ be the group $S_3$. Find the permutation representation of $S_3$. (Note: This gives an isomorphism of $S_3$ into $S_6$.)

I don't know how to write a solution for this problem.
Is the following solution ok?

$$S_3=\{e,\phi,\psi,\psi^2,\phi\psi,\psi\phi\},\text{ }\phi^2=e,\psi^3=e,\phi\psi=\psi^{-1}\phi.$$
$$\tau_e=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\end{pmatrix}.$$
$$\tau_\phi=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\\phi&e&\psi\phi&\phi\psi&\psi^2&\psi\end{pmatrix}.$$
$$\tau_\psi=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\\psi&\phi\psi&\psi^2&e&\psi\phi&\phi\end{pmatrix}.$$
$$\tau_{\psi^2}=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\\psi^2&\psi\phi&e&\psi&\phi&\phi\psi\end{pmatrix}.$$
$$\tau_{\phi\psi}=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\\phi\psi&\psi&\phi&\psi\phi&e&\psi^2\end{pmatrix}.$$
$$\tau_{\psi\phi}=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\\psi\phi&\psi^2&\phi\psi&\phi&\psi&e\end{pmatrix}.$$
The permutation representation of $S_3$ is $\{\tau_e,\tau_\phi,\tau_\psi,\tau_{\psi^2},\tau_{\phi\psi},\tau_{\psi\phi}\}$.

If my solution is ok, then I have a next question:
If we get the Cayley Table of a group $G$, then I think we can easily (almost automatically) get the permutation representation of $G$.
Many similar problems are on pp.81-82 in this book.
What is the intent of these problems?
 A: The way you wrote $S_3$ it is in the form of Dihedral group $D_3$.
For $n=3$, $D_3$ is isomorphic to $S_3$. 
Symmetric group $S_n$ can be defined as set of all permutations of a set having n elements or we can say that it is a collection of all one-one and onto functions from a set(n elements) to itself.
The permutation representation of $S_3$ can be given as
$$S_3=\bigg\{
\begin{pmatrix}
1 & 2 &  3 \\
1 & 2 &  3 
\end{pmatrix},\begin{pmatrix}
1 & 2 &  3 \\
1 & 3 &  2 
\end{pmatrix},\begin{pmatrix}
1 & 2 &  3 \\
3 & 2 &  1 
\end{pmatrix},\begin{pmatrix}
1 & 2 &  3 \\
2 & 1 &  3 
\end{pmatrix},\begin{pmatrix}
1 & 2 &  3 \\
2 & 3 &  1 
\end{pmatrix},\begin{pmatrix}
1 & 2 &  3 \\
3 & 1 &  2 
\end{pmatrix}\bigg\}$$
This is a permutation representation of $S_3$
Now, There exist a subgroup of $S_6$ which is isomorphic to $S_3$. Let us denote that subgroup of $S_6$ by H then
$$H=\bigg\{
\begin{pmatrix}
1 & 2 &  3 & 4 &5 &6 \\
1 & 2 &  3 & 4 &5 &6
\end{pmatrix},\begin{pmatrix}
1 & 2 &  3 & 4 &5 &6 \\
1 & 3 &  2 & 4 &5 &6
\end{pmatrix},\begin{pmatrix}
1 & 2 &  3 & 4 &5 &6 \\
3 & 2 &  1 & 4 &5 &6
\end{pmatrix},\begin{pmatrix}
1 & 2 &  3 & 4 &5 &6\\
2 & 1 &  3 & 4 &5 &6
\end{pmatrix},\begin{pmatrix}
1 & 2 &  3 & 4 &5 &6 \\
2 & 3 &  1 & 4 &5 &6
\end{pmatrix},\begin{pmatrix}
1 & 2 &  3 & 4 &5 &6\\
3 & 1 &  2 & 4 &5 &6
\end{pmatrix}\bigg\}$$
Here $S_3$ is isomorphic to H.
