Dirac group representation I am currently taking a representation theory class (from a physicist), and I am very confused about the Dirac groups' irreducible representations. 
First of all, all the Dirac matrices in the representation have trace = 0, so it does not even seem to include a unit matrix. When we talked about representations in class before, we always had a unit matrix in a representation, what happened? 
Also, the lecture went into distinguishing the case for 2n-dimension and 2n-1-dimension. While I understand why there is one more conjugacy classes in the odd dimension (thus even dimension having one more irreducible rep than odd dimension), I can't fully appreciate all the difference in the irreducible representations in the cases of even and odd dimensions; in particular, I was asked in a homework to show that if a Dirac matrices {$\gamma^\mu$} form an irreducible rep, then show that {$-\gamma^\mu$} is equivalent irreducible rep in the case of even dimension, and inequivalent irreducible rep in the case of odd dimension. But then again, if I think about the character table to see whether a representation is equivalent or inequivalent to another representation, I feel like matrices in {$\gamma^\mu$} and {-$\gamma^\mu$} will never have same trace, thus they can never be equivalent (unless they are all 0, which is the case, I believe. Then again, how could they be different representations then?).
I would appreciate any good reading materials / answers to my questions!
 A: The defining relation for the Clifford algebra, $Cl(1,d)$ is
$$
\{\gamma_\mu,\gamma_\nu\}=2 \eta_{\mu\nu}\ \mathbf{1}\ ,
$$
For simplicity, I will assume that $\eta_{\mu\nu}=\text{Diag}(1,-1,\ldots,-1)$ with $\mu,\nu=0,1,\ldots,d$. Other signatures can easily be incorporated. It is easy to see that $\gamma_0^2=-\gamma_i^2=\mathbf{1}$ for $i=1,\ldots,d$. Using the defining relation, one has
$$
\gamma_0 \gamma_i + \gamma_i  \gamma_0 =0 \ .
$$
Multiply the above equation by $\gamma_0$ and then take the trace to obtain
$$
\text{Tr}(\gamma_i) + \text{Tr}(\gamma_0 \gamma_i \gamma_0)=0\implies \text{Tr}(\gamma_i)=0\ ,
$$
on using the cyclic property of the trace. Similarly, one can show $\text{Tr}(\gamma_0)=0$. So the defining property proves the tracelessness of the Dirac matrices.
Two representations, $\gamma_\mu$ and $\gamma_\mu'$, of the Clifford algebra are said to be equivalent if $\gamma_\mu' = S \cdot \gamma_\mu S^{-1}$ for some invertible matrix $S$.
Appendix A of the Physics Reports article by Sohnius might be a good starting point for the other properties.
A: Although the responses so far have been illuminating, I believe they missed your main point of confusion (the first commenter nailed it I think). 
One must differentiate between the Clifford algebra $\text{Cℓ}_{t,s}(\mathbb{R})$, where $t+s=d$ is the dimensionality of spacetime, and the Dirac matrices which generate a basis for it via products. More specifically, we have a set of $d$ Dirac matrices $\{\gamma^\mu\}$ which satisfy the Clifford algebra relation $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$. Products of these Dirac matrices, along with the identity, form a basis for the Clifford algebra:
$$\Gamma = \{1,\gamma^\mu,\gamma^5,\gamma^\mu\gamma^5,\gamma^\mu \gamma^\nu\}$$
where $\gamma^5=\prod_\mu \gamma^\mu$, with a potential factor of $i$ to ensure Hermiticity depending on $(t,s)$.
This common choice of basis for the Clifford algebra forms a finite group$^\dagger$, and is what people refer to as the "Dirac group" - not the individual Dirac matrices themselves.
Indeed, as you noticed (and as @suresh confirmed), the unit matrix cannot be a Dirac matrix in any representation (since $\text{Tr}(\gamma^\mu )=0$ for all $\mu$). But the unit matrix has to be present in any representation of any finite group!

$^\dagger$ Actually, as I've written it $\Gamma$ does not generally form a group. One must prepend $\pm$ to each entry: $\Gamma_{\text{Dirac}}=\{\pm 1, \pm \gamma^\mu, \pm\gamma^5, \pm\gamma^\mu\gamma^5, \pm\gamma^\mu \gamma^\nu\}$
