Does a group (finite or infinite) always be isomorphic to a subgroup of $GL_n(\mathbb C)$? Let $G$ be an arbitrary group (finite or infinite), and let $GL_n(\mathbb C)$ be the general liner group. And let $\varphi : G \rightarrow GL_n(\mathbb C)$ be a homomorphic map. 
My question is: Can we always find the homomorphic map $\varphi$ be injective? In another word, can any infinite group be represented by some $n \times n$ complex matrices, which are matched with a single element in $G$?
Above are some examples.
Finite group like Hamilton IV group, $\mathrm Q=\{\mathrm{1, i, j, k}\}$, can be represented by these 8 complex matrices, where:
$1=\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},\mathrm{i}=
\begin{pmatrix}
i & 0 \\
0 & -i
\end{pmatrix},\mathrm{j}= 
\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix},\mathrm{k}=
\begin{pmatrix}
0 & i \\
i & 0
\end{pmatrix}.
$
And infinite group like $\mathbb Z^*$ can be simply represented by $1\times 1$ matrices:
$(e^n), n \in \mathbb Z.$
So this problem occured to me.


Now, one idea is that the infinite group $G$ is generated by finite elements $\{1, x_1, ..., x_n\}$, and these elements are somewhat 'isolated' from each other. The word 'isolated' may mean that $ x_m \not= x_1^{r_1}x_2^{r_2}\dots x_{m-1}^{r_{m-1}}.$ But at least now, there are still 3 bad things about the idea: 
 1. How to define the word 'isolated'?
 2. How to solve the problem of non-abelian situations?
 3. Some infinite groups such as $\mathbb R^*$ cannot be generated by a few elements, but it can still be represented by matrices.
 A: All finite groups are isomorphic to a subgroup of $Gl_n(\mathbb{C})$ for some $n$. This can be shown by first showing that all finite groups are subgroups of a permutation group and then that all permutation groups sit inside a matrix group.
Infinite groups in general need not be isomorphic to a subgroup of $Gl_n(\mathbb{C})$ for any $n$. If I recall correctly one of the smallest/ simplest counter examples consists of infinitely many copies of $\mathbb{Z}_2$. Essentially one can build infinite groups that are just too big to fit into $Gl_n(\mathbb{C})$.
A: I'll write a stronger statement: if $G$ is an abelian group of finite exponent (so all the elements have order less than, say, $r$), then $G$ admits an embedding in some $GL_n(\mathbb{C})$ if and only if it is finite.
Indeed, if $G$ is finite, it has a faithful representation, like any finite group. And the issue when $G$ is infinite is that $GL_n(\mathbb{C})$ cannot have an abelian subgroup of exponent $r$ with more than $r^n$ elements, because such a subgroup can be simultaneously diagonalized, and is therefore, up to conjugation, a subgroup of $\mu_r(\mathbb{C})^n\subset D_n(\mathbb{C})\subset GL_n(\mathbb{C})$ (here $D_n(\mathbb{C})$ is the subgroup of diagonal matrices in $GL_n(\mathbb{C})$).
