Can every group be represented by a group of matrices? 
Can every group be represented by a group of matrices?

Or are there any counterexamples? Is it possible to prove this from the group axioms?

For finite groups, the answer is yes:  every finite group is isomorphic to a subgroup of $Gl(n)$ for some $n$ large enough.  In fact, we can do slightly better and embed it into $SO(n)$ for $n$ large enough.
The idea:  Every group acts on itself, by, say, left translations.  This gives an embedding $G\rightarrow S_{|G|}$ into the symmetric group on $|G|$ letters.  Hence, every group is a subgroup of some $S_k$ for $k$ large enough.
Thus, we just need to show that we can embed every $S_n$ into $GL_n$ for $n$ large enough.  But $S_n$ acts on $\mathbb{R}^{n}$ by permuting the coordinates.  It's easy to see that this action is linear, hence defines an embedding $S_n\rightarrow GL(n)$.
If you insist on positive determinant, one can include $GL(n)$ into $Gl(n+1)$ sending a matrix $A$ to $\operatorname{diag}(A, \operatorname{sign}(\det(A)))$.
Finally, to embed orthogonally, pick your favorite inner product $\langle, \rangle$ on $\mathbb{R}^n$ and define $\langle x, y\rangle_1 = \sum_{g\in G} \langle g(x),g(y)\rangle$.  It's easy to see that this new metric is $G$ invariant, so $G$ acts by isometries in this metric, hence, $G$ embeds into $SO(n)$ for $n$ large enough.
The story for infinite groups is more complicated.  Of course, there are groups of any preassigned cardinality, while any subset of $Gl(n)$ can have cardinality at most that of the real numbers.  So, no sufficiently large group can embed into $Gl(n)$.
If you mix a bit of topology, then, for example, every compact Lie group embeds into $SO(n)$ for $n$ large enough, but the proof is not trivial.  Noncompact Lie groups need not embed into $GL_(n)$ for any $n$.  For example, I believe that the universal cover of $Sl(2)$ is such an example.
