Basis for a tensor product of two modules over a group algebra Let $G$ be a finite group and let $H \subset G$ be a subgroup of index $n$. Let $K$ be a field and let $V$ a vector space of dimension $m$ over $K$. Suppose $\rho : H \rightarrow \text{GL}(V)$ is a representation of $H$ on $V$. We can then regard $V$ and the group algebra $K[G]$ as $K[H]$-modules. Hence we can form the $K[H]$-module $K[G] \otimes_{K[H]} V$ which by restricting scalars is also $K$-vector space. 
Suppose $g_1, \dots, g_n$ are elements of $G$ representing $G/H$ and suppose the vectors $e_1, \dots, e_m \in V$ form a $K$-basis of $V$. I would like to prove that the set $\{g_i \otimes e_j \,:\, 1 \leq i \leq n, 1 \leq j \leq m\}$  is a $K$-basis of $K[G] \otimes_{K[H]} V$.
I have already proved that it generates $K[G] \otimes_{K[H]} V$ over $K$. 
Now suppose that we have scalars $t_{ij} \in K$ such that
$$\sum_{i=1}^n\sum_{j=1}^m{t_{ij}(g_i \otimes e_j)} = 0$$
Defining $v_i :=\sum_{j=1}^m{t_{ij}e_j}$ we can rewrite this as
$$\sum_{i=1}^n{(g_i \otimes v_i)} = 0$$
I would like to deduce from this that each $v_i$ must be zero. Is this possible or is there a different approach needed?
Thanks
 A: For every $i = 1, \dotsc, n$ let
$$
     K[H]_i
  := \bigoplus_{g \in g_i H} K g
   = \bigoplus_{h \in H} K (g_i h).
$$
(We may think of $K[G]_i$ as $K[g_i H]$.)
We have a decomposition of right $H$-sets given by
$$
  G = \coprod_{i=1}^n g_i H.
$$
This linearizes to a decomposition of right $K[H]$-modules given by
$$
  K[G] = \bigoplus_{i=1}^n K[H]_i.
$$
For each $g_i H$ we have an isomorphism of right $H$-sets
$$
  H \to g_i H,
  \quad
  h \mapsto g_i h
  \quad
  \text{for all $h \in H$}
$$
which linearizes to an isomorphism of right $K[H]$-modules
$$
  K[H] \to K[H]_i,
  \quad
  h \mapsto g_i h
  \quad
  \text{for all $h \in H$}.
$$
With this we find that
$$
    K[G]
  = \bigoplus_{i=1}^n K[H]_i
  \cong \bigoplus_{i=1}^n K[H]
$$
is just a free right $K[H]$-module, and we get the following isomorphism of $K$-vector spaces:
$$
        K[G] \otimes_{K[H]} V
  =     \bigoplus_{i=1}^n K[H]_i \otimes_{K[H]} V
  \cong \bigoplus_{i=1}^n K[H] \otimes_{K[H]} V
  \cong \bigoplus_{i=1}^n V
$$
By pulling back the basis of $\bigoplus_{i=1}^n V$ which is induced by the  given basis of $V$, we arrive at the claimed basis of $K[G] \otimes_{K[H]} V$.
