Basis of induced representation

Question

Let $G$ be a group and $H$ a subgroup. My lecture book (Algebra by Cohn) defines the (right) $G$-module induced by a right $H$-module $U$ by $$\tag{1} \text{ind}_H^GU=U\otimes_{kH} kG, $$ where $kH$ and $kG$ are the group algebras of $H$ and $G$, respectively. Since the tensor product is balanced with respect to the elements of $kH$, the basis of $(1)$ must be given by $$\tag{2} B_1=\{\mathbf{u}_i\otimes_{kH}\mathbf{e}_g\mid \mathbf{u}_i\in B_U, \mathbf{e}_g\notin kH\setminus \{1\}\}, $$ where $B_U$ is a basis of $U$ and $\mathbf{e}_g$ a basis element of $kG$. However, Cohn writes that the basis can be written $$\tag{3} B_2=\{\mathbf{u}_i\otimes_{kH}\mathbf{e}_t\mid \mathbf{u}_i\in B_U, t\in T\} $$ where $T$ is a transversal for $H$ in $G$. But this does not seem to add up, because $\dim{B_1}=\dim{U}(|G|-|H|+1)$, whereas $\dim{B_2}=\dim{U}|T|=\dim{U}|G|/|H|$, and these values are not necessarily the same, hence, either $B_1$ or $B_2$ is not a basis. I assume that the author is correct, so $B_1$ is not a basis, but why?

Answer 1

so $B_1$ is not a basis, but why?

It is linearly dependent if $H$ is not trivial.

For instance, for any $g\in G\setminus H$, $h\in H$ and basis vector $u_i$ we have $u_i\otimes hg=u_ih\otimes g$, and $u_ih$ can be written as a linear combination of basis vectors. You avoided producing linear dependence when $g\in H$ by forcing only one member of $H$ to be used in the basis (the identity $1$), but this logic needs to be applied not just in $H$ itself but in every coset of $H$.