Finite Subgroups of $GL(n,\mathbb{C})$ In the Artin's book on Algebra, the author stated a theorem (Ch.9, Thm. 2.2):
"A finite subgroup $G$ of $GL(n,\mathbb{C})$ is conjugate to a subgroup of $U(n)$. 
Here, $U(n)$ is the unitary group, i.e. if $\langle \,\,, \rangle$ is the standard Hermitian inner product on $\mathbb{C}^n$ given by 
$\langle (x_1,\cdots,x_n),(y_1,\cdots,y_n)\rangle=\sum_{}{x_i\bar{y_i}}$ then $U(n)=\{A\in GL(n,\mathbb{C})  \colon \langle Av,Aw\rangle = \langle v,w\rangle,\forall v,w\in \mathbb{C}^n \}$. 
The proof is: there is a $G$-invariant Hermitian inner product $\langle\,\, , \rangle_1$ on $\mathbb{C}^n$, and consider an orthonormal basis $B_1$ w.r.t this form. If $P$ is the matrix of transformation which changes standard basis to $B_1$, then $PGP^{-1}\leq U(n)$
Question: In the last statement in the proof, the author says that $PGP^{-1}$ is subgroup of unitary group; but, this unitary group is corresponding to the Hermitian inner product $\langle \,\,, \rangle_1$, i.e. it is the group of linear transformations which preserves the inner product $\langle \,\,, \rangle_1$. Why $PGP^{-1}$ should be subgroup of $U(n)$ which is the group of linear transformations which "preserves" the standard Hermitian inner product $\langle \,\,,\rangle$ on $\mathbb{C}^n$?
 A: Preserving the $G$-invariant inner product $\langle , \rangle_1$ implies preserving the original $\langle , \rangle .$ To see this requires knowing how $\langle , \rangle_1$ was constructed (via Weyl's averaging trick.) For $u,v \in \mathbb{C}^n,$ $g\in G,$ and $A \in PGP^{-1},$ we have  
$\langle A(gu), A(gv) \rangle_1 = \langle gu, gv \rangle_1 =\langle u, v \rangle_1 =\frac{1}{|G|}   \displaystyle\sum_{g\in G} \langle u , v \rangle = \langle u, v \rangle .$ 
A: To answer the question a slightly different way: choose an orthonormal basis wrt the $G$-invariant Hermitian form $\langle, \rangle_{1}.$ Let this be $\{ u_{i}: 1 \leq i \leq n \}.$
Since the form is $G$-invariant, for any fixed $g \in G,$ 
we have $\langle gu_{i},gu_{j} \rangle_{1} = \delta_{ij}$ for all $i,j.$ Hence writing the matrix of (the transformation represented by) $g$ with respect to this basis, the columns (and rows) are orthonormal, and the matrix is unitary. So now you have a group of unitary matrices, which is conjugate to the original one via a change of basis matrix.
To answer comment below, which took to long to leave as a comment: It means that if write $gu_{i} = \sum_{j=1}^{n} a_{ij}(g)u_{j},$ where the $a_{ij}(g)$ are complex numbers, then we have for any $r,s$, $$\delta_{rs} = \langle gu_{r},gu_{s} \rangle_{1} = \sum \sum_{p,q= 1}^{n} a_{rp}(g) {\overline a_{sq}(g)} \langle u_{p}, u_{q} \rangle_{1} .$$ But this is $$\sum_{p =1}^{n} a_{rp}(g) { \overline a_{sp}(g)}$$    because $\langle u_{p}, u_{q} \rangle_{1} = \delta_{pq}.$  So what is left is the usual innner product of the rows of the matrix  $[a_{ij}(g)].$ 
