Are all compact subgroups of $GL(n,\Bbb C)$ in $U(n)$? If $G$ is a compact subgroup of the multiplicative group $\Bbb C-\{0\}$, then it is easy to show that $G\subseteq S^1$. I wonder if this generalizes as follows:

Question: If $G$ is a compact subgroup of $GL(n,\Bbb C)$, do we have $G\subseteq U(n)$?

I am more interested in Lie groups, but I don't know if assuming $G$ is a Lie group can help. If it does make a difference, then yes I am assuming $G$ is a Lie group.

Proof in case of $n=1$: Suppose $g\in G$ and $g\notin S^1$. Then, $g=re^{i\theta}$ for $r\neq 1$. By taking $g^{-1}$ if necessary we may assume $r>1$. Then, $g^n\in G$ for all $n\geq 0$ but $|g^n|=r^n\to\infty$ as $n\to\infty$ so $G$ is unbounded and hence not compact.

 A: No it is not true. For $n=2$, define $A\in M_2(\mathbb{C})$ :
$$A:=\begin{pmatrix}1&-2\\0&-1\end{pmatrix} $$
We have that :
$$A^2=\begin{pmatrix}1&-2\\0&-1\end{pmatrix}\begin{pmatrix}1&-2\\0&-1\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix} $$
Hence $G:=\langle A\rangle$ is a group of order $2$, in particular, it is compact. But clearly $AA^*\neq I_2$ so $G$ is not included in $U(2)$. 
Remark 1 : $A$ is the involution taking $e_1$ to $e_1$ and $e_1+e_2$ to $-(e_1+e_2)$ (where $(e_1,e_2)$ is an orthonormal base for the natural hermitian inner-product of $\mathbb{C}^2$). I constructed it as a natural symmetry for a non-hermitian base, it is not so surprising that this leads to a non-hermitian transformation.
Remark 2 : conjugating $A$ by the changement of base $(e_1,e_2)$ to $(e_1,e_1+e_2)$, $A$ becomes $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ which is an hermitian transformation. Hence up to conjugation $A$ is included in $U(2)$. 
Remark 3 : In general, any finite group $G$ of $GL(n,\mathbb{C})$ is included in $U(n)$ up to conjugation. The proof :

 Let $\langle\cdot,\cdot\rangle$ be the hermitian inner product on $\mathbb{C}^n$. Define a new hermitian inner product on $\mathbb{C}^n$ : $$\langle u,v\rangle_{G}:=\frac{1}{|G|}\sum_{g\in G}\langle g\cdot u,g\cdot v\rangle$$ One should check that $\langle \cdot,\cdot\rangle_{G}$ is indeed an hermitian inner product. Once you have these you can also realise that any $g\in G$ is a hermitian transformation for $\langle \cdot,\cdot\rangle_{G}$. Hence $G\leq U(\langle \cdot,\cdot\rangle_{G})$. Finally the groups $U(\langle \cdot,\cdot\rangle_{G})$ and $U(n)$ are conjugate because all inner product admits an orthonormal base : once an orthonormal base $\beta$ for $\langle \cdot,\cdot\rangle$ and an orthonormal base $\beta_G$ for $\langle \cdot,\cdot\rangle_G$ are given then the matrix sending the base $\beta$ to $\beta_G$ clearly conjugates $U(n)$ to $U(\langle \cdot,\cdot\rangle_{G})$.

Remark 4 : the same is true for $G$ compact subgroup of $GL(n,\mathbb{C})$. The "proof" :

 By the proof for remark 3, it suffices to find a hermitian inner product stable by $G$. It is given by changing the sum to an integral on $G$ for a Haar-mesure. Since $G$ is compact, it is well-defined. The invariance of the inner-product is using the fact that we choose a Haar-mesure which is (by definition) right-invariant. 

