The bicyclic semigroup $B$ is the semigroup with two generators $p, \; q$ and the single relation that $pq = 1$. So all other words in $B$ are of the form $q^{n}p^{m}$ for $m, n \in \mathbb{N}$.
I would like to know if there are any faithful representations of $B$. I don't think there can be any such finite dimensional (matrix) representations because if $\rho: B \rightarrow GL(\mathbb{C}^{n})$ is such a representation, then there are no finite dimensional matrices such that $\rho(p)\rho(q) = I \neq \rho(q)\rho(p)$, as the bicyclic semigroup would require. So... what are the representations of the bicyclic semigroup? Are there faithful ones?
As Izaak says in the comments, there are plenty of finite-dimensional representations (given by any invertible matrix), just not faithful ones. The finite-dimensional representations are all representations of $\mathbb{Z}$, so the irreducible ones are $1$-dimensional and given by taking $\rho(p)$ to be multiplication by a nonzero scalar $\lambda$ and taking $\rho(q)$ to be multiplication by $\lambda^{-1}$.
A natural infinite-dimensional faithful representation is to take $V = \mathbb{C}^{<\mathbb{N}}$ to be the vector space of infinite sequences with finite support, to take $\rho(q)$ to be the right shift
$$\rho(q) : (c_0, c_1, \dots) \mapsto (0, c_0, c_1, \dots )$$
and to take $\rho(p)$ to be the left shift
$$\rho(p) : (c_0, c_1, \dots ) \mapsto (c_1, c_2, \dots ).$$
Another way to think about this representation is that $V$ can be identified with the vector space $\mathbb{C}[x]$ of polynomials in one variable (via its monomial basis, so we identify $(c_0, c_1, \dots)$ with the polynomial $f(x) = \sum c_i x^i$). Then the right shift $\rho(q)$ can be identified with the multiplication operator $f(x) \mapsto xf(x)$, while the left shift $\rho(p)$ can be identified with the operator $f(x) \mapsto \frac{f(x) - f(0)}{x}$. Either way it's not hard to show that $\rho(pq) = 1$ and that the operators $\rho(q^a p^b)$ are distinct, and I think even linearly independent.
This representation $V$ is also irreducible. To see this it suffices to show that any nonzero vector generates. Just apply the left shift until we get a vector of the form $(c, 0, 0, \dots)$, then repeatedly take right shifts and linear combinations to get every other vector.
In general, every monoid $M$ whatsoever has a faithful representation of dimension $|M|$, given by taking the free vector space $\mathbb{C}[M]$ on the left action of $M$ on itself. This is the regular representation, and $M$ can be replaced by any other set on which $M$ acts faithfully. $V$ can be thought of as being obtained by starting from the regular representation of $B$ and quotienting by the subrepresentation generated by $p$. This has the effect of leaving only powers of $q$ and then to identify this with the polynomial representation above we identify $\sum c_i x^i$ with $\sum c_i q^i$.