Computing regular representations I'm doing a reading of representation theory (not a class) but don't really understand what $GL(\mathbb{F}(G))$ is. We defined $GL(n,\mathbb{C})$ in the usual way (set of matrices with nonzero determinant). But I'm not sure the determinant has a natural generalization for free vector spaces (ie $\mathbb{F}(G)$). Also, because I am not sure about this or what the objects look like in this space, I can't answer this fairly easy question (help appreciated here): Let $\mathbb{Z}_4$ denote the cyclic group of order four, compute the four by four matrices of each of the four elements for the left and right representations.  Actually, I would only like one matrix and a brief explanation of how it was found because I'm certain I can do the rest.  I suspect the matrix for $0$ would be the identity and the matrix for $1 $would have for rows $e_4, e_1, e_2,e_3$ respectively... but I'm not sure precisely why.... because I don't understand what $GL(\mathbb{F}(G))$ is.
 A: If I understand the notation correctly, ${\mathbb{F}(G)}$ is simply the free vector space generated by $G$. Explicitly, ${\mathbb{F}(G)}$ is the vector space spanned by elements $[g]$ for $g\in G$: Elements of it are given by sums $\alpha_i[g_1] + \cdots + \alpha_n [g_n]$ for $g_i\in G$ distinct; addition is given by
$$\sum_{g\in G} \alpha_g [g] + \sum_{g\in G} \beta_g [g] = \sum_{g\in G} (\alpha_g + \beta_g)[g];$$
and the action of ${\mathbb{F}}$ is given by $k.[g] = k[g]$. (I'm assuming $G$ is finite here, for simplicity.) Then $\mathbb{F}(G)$ has dimension $\#G$, so its determinant of a homomorphism $\mathbb{F}(G) \to \mathbb{F}(G)$ is defined as for any other vector space over $\mathbb{F}$. (If you're familiar with algebraic number theory, the definition of the norm for a field extension $L/K$ is very similar to the construction here.)
Here's an explicit example. Take $G = {\mathbb{Z}}_4$, and write $G = \{1, x, x^2, x^3\}$ for some generator $x\in G$. We'll use that as our basis of $V = \mathbb{F}(G)$: the vectors $[1], [x], [x^2],$ and $[x^3]$. The left action of $G$ on itself, given by $\rho(g)h = gh$, defines the left representation of $G$: $\rho(g)[h] = [gh]$, and we extend $\rho$ by linearity to get a map $\rho:G \to \text{End}(V)$. This map is invertible, since every element of $G$ is invertible, so we actually get a map $\rho: G \to GL(V)$. For example, $\rho(x)$ is the map that takes $[1]\to [x]$, $[x]\to [x^2]$, $[x^2]\to [x^3]$, and $[x^3] \to [1]$. Over the basis, we've chosen, we wind up with:
$$\rho(x) = \begin{pmatrix}0 & 0 & 0 & 1\\1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\end{pmatrix}.$$
Again, this is just the matrix form of an ordinary endomorphism of a 4-dimensional vector space over ${\mathbb{F}}$, and its determinant is entirely analogous to what it would be if the matrix above were just an ordinary element of $GL_4(\mathbb{C})$. You can derive the other $\rho(x^i)$ similarly, or use the fact that $\rho(x^i) = \rho(x)^i$. 
In short: ${\mathbb{F}(G)}$ is the vector space spanned by symbols $[g]$ for $g\in G$. (This is just formal notation; I'm not even using the fact that $G$ is a group yet.) The left representation $\rho$ is defined by (linearity and the relation) $\rho(g)[h] = [gh]$. Choose your favorite basis of ${\mathbb{F}(G)}$ (the elements of $G$ listed in a particular order is generally a good choice), and you can write down $\rho$ as a matrix.
