Consider a group $G$ and a representation $\rho=(\rho_{ij})$ of degree $d$ as well as a $d$-dimensional vector space $V$. In the book Algebra by P. M. Cohn he states that we can turn $V$ into a $G$-module by defining the action of $x\in G$ on a basis $v_1,...,v_d$ by $$ x v_i=\sum_j \rho_{i j}(x) v_j \quad(x \in G) $$ and generally putting $$ x \left(\sum \alpha_i v_i\right)=\sum \alpha_i \rho_{i j}(x) v_j. $$ What I understand from this is that we define the action of $x$ on a linear combination of the basis vectors by $$ x\left(\sum_i \alpha_i v_i\right)=\sum_i \alpha x v_i=\sum_i \alpha_i \sum_j \rho_{i j}(x) v_j=\sum_{i, j} \alpha_i \rho_{i j}(x) v_j. $$ (Correct?)
He leaves it to the reader to show that it is a $G$-module, so I need to show that the multiplication satisfies for any $x,y\in G$, $\boldsymbol{v},\boldsymbol{w}\in V$ and $c,d$ in the field:
- $x \boldsymbol{v} \in V$,
- $x(c \boldsymbol{v}+d \boldsymbol{w})=c(x \boldsymbol{v})+d(x \boldsymbol{w})$
- $(x y) \boldsymbol{v}=x(y\boldsymbol{v})$
- $\epsilon\boldsymbol{v}=\boldsymbol{v}$
My intend:
The linear combination of the basis vectors $\sum \alpha_i \rho_{i j}(x) v_j$ is in $V$ by definition of a vector space, so $x \boldsymbol{v} \in V$.
Let $\boldsymbol{v}=\sum_i \alpha_i v_i$ and $\boldsymbol{w}=\sum_i \beta_i v_i$ be arbitrary vectors in $V$. Then $$ \begin{aligned} x(c \boldsymbol{v}+d \boldsymbol{w}) & =x\left(\sum_i\left(c \alpha_i+d \beta_i\right) v_i\right) \\ & =\sum_i\left(c \alpha_i+d \beta_i\right) \rho_{i j}(x) v_j \\ & =\sum_i\left(c \alpha_i \rho_{i j}(x) v_j+d \beta_i \rho_{i j}(x) v_j\right) \\ & =c \sum_i \alpha_i \rho_{i j}(x) v_j+d \sum_i \beta_i \rho_{i j}(x) v_j \\ & =c x \boldsymbol{v}+d x \boldsymbol{v} \end{aligned} $$
We have $$ (x y)\left(\sum_i \alpha_i v_i\right)=\sum_{i, j} \alpha_i \rho_{i j}(x y) v_j=\sum_{i, j, k} \alpha_i \rho_{i k}(x) \rho_{k j}(y) v_j $$ and $$ x\left(y \sum_i \alpha_i v_i\right)=x \sum_{i, k} \alpha_i \rho_{i k}(y) v_k=\sum_{i, k} \alpha_i \rho_{i k}(y) x v_k=\sum_{i, j, k} \alpha_i \rho_{i k}(y) \rho_{k j}(x) v_j $$ but this doesn't appear to be the same - what am I doing wrong?
Since $\rho$ is a representation it sends the identity $\epsilon\in G$ to the identity matrix and hence $$ \epsilon \boldsymbol{v}=\sum_i \alpha_i \rho_{i j}(\epsilon) v_j=\sum_i \alpha_i \delta_{i j} v_j=\boldsymbol{v} $$ Can someone see if my proofs are correct, and let me know what I am missing/misunderstanding in part 3 (associativity).