TITLE: Presentation of the Clifford group by generators and relations?
QUESTION [6 upvotes]: The Clifford group $\mathcal{C}_n$ is a matrix group on $\mathbb{C}^{2^n}$ generated by tensor products of the following matrices:
$$
P = \begin{pmatrix} 1 & 0 \\\\ 0 & i\end{pmatrix}
\quad
H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\\\ 1 & -1\end{pmatrix}
\quad
\wedge\\! X = \begin{pmatrix}1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0\end{pmatrix}
$$
For my purposes I take equality up to a complex unit, i.e., if $A = e^{i\alpha}B$ then $A = B$.
My question is: 

is there an abstract presentation of this group by generators and relations?

Or the easier version: 

is there a presentation of $\mathcal{C}_2$ or $\mathcal{C}_3$ by generators and relations?

I don't care which set of generators is used, although it would be nice if the generators were those given above.

REPLY [4 votes]: Follow-up, many years later:
In the currently (02/2020) most highly up-voted comment and the discussion below it, I claimed that the Clifford group is a projective representation of the semi-direct product of the phase space translations and the symplectic group. As Ross correctly implies in his response to my answer (and I wrongly deny in my response to him), this property only holds for qudits in odd dimensions, not for the qubit case! For qubits, it is still true that the symplectic group arises as a quotient group of the Clifford group (up to Paulis and multiples of the identity). However, the extension does not split.
[I've since lost my old account and access to the old email address - hence this answer comes from a new account. Apparently, the new account does not have enough "reputation" to directly reply.]