I solved the following exercise, could someone please check my work?
Exercise: Let
$$ A = \left ( \begin{array}{cccc} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right ) $$
(1) Calculate $R_A(x,y,z,w)$.
(2) Describe a subgroup $H$ of $O(4)$ which is isomorphic to $S_4$ ($S_4$ is the group of permutations of a $4$-element set)
(3) Describe a subgroup $H$ of $O(n)$ which is isomorphic to $S_n$. What is $H \cap SO(n)$?
(4) Prove that every finite group is isomorphic to a subgroup of $O(n)$ for some integer $n$.
My work:
(1) $R_A (x,y,z,w) = (y, z, w, x)$
(2) The subgroup generated by all permutations of columns of $A$. It is clear that a permutation of the columns corresponds to a permutation of the elements in the vector $(x,y,z,w)$.
(3) Let $H$ be the group generated by all even permutations of the identity matrix. Since $H \subseteq SO(n)$ we have $H \cap SO(n) = H$.
(4) By Cayley's theorem every group is isomorphic to a group of permutations and by (3), this group is isomorphic to a subgroup of $O(n)$.
If you take all matrices that are formed by permuting the columns of an $n\times n$ identity matrix, you get a group of matrices isomorphic to $S_n$.
The isomorphism can be defined as follows
$$i:S_n\to H : \pi \mapsto i(\pi) \text{ with } (i(\pi))_{kl}=\delta_{k\pi(l)}$$
in which $\delta_{kl}$ is the Kronecker delta.
It is clear that it is bijective, and also note that given two permutations $\pi$ and $\rho$
$$(i(\pi)\cdot i(\rho))_{km} = \sum_{l=1}^n i(\pi)_{kl}\cdot i(\rho)_{lm} = \sum_{l=1}^n \delta_{k\pi(l)}\cdot \delta_{l\rho(m)}=\sum_{l=1}^n \delta_{\pi^{-1}(k)l}\cdot \delta_{l\rho(m)}=\delta_{\pi^{-1}(k)\rho(m)}=\delta_{k(\pi\circ\rho)(m)} = (i(\pi \circ \rho))_{km} \; .$$
Now $H\cap SO(n)$ implies that we add the additional constraint that the determinant be $1$. Odd permutations of the identity matrix will have determinant $-1$, so those are out. We are left with even permutations, but per our isomorphism, we know that those matrices will form a group isomorphic to $A_n$. Thus $H\cap SO(n) \cong A_n$.
Note that your answer is wrong because in (3) they explicitly define $H$ as being the subgroup of $O(n)$ isomorphic to $S_n$ and then ask about $H\cap SO(n)$. But then you go and redefine $H$ to be the group of all even permutations. You can't just do that.
By Cayley's theorem, every group is isomorphic to a group of permutations, but in particular every finite group is isomorphic to a finite group of permutations. And since we just showed that every finite group of permutations is isomorphic to a subgroup of $O(n)$, we obtain (4). Note the emphasis on finite is important and was lacking in your answer.