Show that every finite group of order $n$ is isomorphic to a group consisting of $n\times n$ permutation matrices. I solved the problem myself and I'm curious if this is legitimate.
I used that if a group of order $n$, $X$, is isomorphic to a subgroup of $A$,
$X$ should be also isomorphic to a subgroup of $B$ if $A$ and $B$ is isomorphic.
(because I think this is an isomorphic structure?)
Please help me if there is any error.
Thank you!

A permutation matrix is one that can be obtained from an identity matrix by reordering its rows. If $P$ is an $n\times n $ permutation matrix and $A$ is any $n\times n$ matrix and $C=PA$, then $C$ can be obtained from $A$ by making precisely the same reordering of the rows of $A$ as the reordering of the rows which produced $P$ from $I_n$.
$\textbf{a.}\,\,$ Show that every finite group of order $n$ is isomorphic to a group consisting of $n\times n$ $\quad\,\,\,$permutation matrices undert matrix multiplication.
$\textbf{b.}\,\,$ For each of the four elements $e,a,b$, and $c$ in the Table $5.11$ for the group $V$, give a $\quad\,\,\,$specific $4\times 4$ matrix that corresponds to it under such an isomorphism.

 A: 
I used that if a group of order $n$, $X$, is isomorphic to a subgroup of $A$, $X$ should be also isomorphic to a subgroup of $B$ if $A$ and $B$ is isomorphic.

Yes. "$X$ is isomorphic to a subgroup of $A$" is the same as saying "there is an injective homomorphism $f: X\to A$". And "$A$ is isomorphic to $B$" is the same as saying "there is a bijective homomorphism $g: A\to B$". It's easy to see that $g\circ f: X\to B$ is an injective homomorphism, i.e. $X$ is isomorphic to a subgroup of $B$. (Actually, $g$ doesn't need to be surjective.) As long as all of your maps are structure-preserving, the structure gets preserved no matter how many maps you hit it with!
You now need to show that $P_n$ is isomorphic to $S_G$, where $G$ is a group of size $n$. Well, $P_n$ is a group of $n\times n$ matrices, so I recommend you pick a basis $e_1, \dots, e_n$ for those matrices; and $S_G$ is the permutation group of a set of size $n$, so I recommend you give the elements of that set names, say $g_1, \dots, g_n$. An element of $P_n$ just swaps around the $e_i$; an element of $S_G$ just swaps around the $g_i$. Can you see how the isomorphism $P_n \to S_G$ works?
