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 $$\tag{1} \text{ind}_H^GU=U\otimes_H kG, $$ where $kG$ is the group algebra. He writes that if we take a coset representation of $G$ $$\tag{2} G=Ht_1\cup\dots \cup Ht_r, $$ then we can write Equation $(1)$ as $$\tag{3} \text{ind}_H^GU=U\otimes Ht_1\oplus \dots\oplus U\otimes Ht_r. $$ However, I am not sure how he gets from Equation $(1)$ to Equation $(3)$. I think that one thing that makes it more confusing is that he uses sloppy notation. The group algebra is defined $$\tag{4} kG:=\Big\{\sum_{g\in G}a_g\boldsymbol{g}\mid a_g\in k\Big\}. $$ Define $$\tag{5} (kH)t_i:=\Big\{\sum_{h\in H}a_h\boldsymbol{ht}_i\mid a_h\in k\Big\}. $$ Since the collection of $t_i$ define a transversal, each $g\in G$ can be written uniquely in the form $ht_i$, so an arbitrary $\sum_{g\in G}a_g\boldsymbol{g}\in kG$ can be written uniquely in the form $$\tag{6} \sum_{g\in G}a_g\boldsymbol{g}=\sum_{h,i}a_{h}^{(t_i)}\boldsymbol{ht_i}=\sum_{i}\Big(\sum_{h}a_{h}^{(t_i)}\boldsymbol{ht_i}\Big). $$ By definition of the direct sum of vector spaces, this implies that $$\tag{7} kG=\bigoplus_{i}(kH)t_i, $$ Since the tensor product is distributive with respect to a direct sum, it then follows from Equation $(7)$ that we can rewrite Equation $(1)$ as $$\tag{8} \text{ind}_H^GU=U\otimes_H kG=\bigoplus_iU\otimes_H (kH)t_i. $$ It therefore appears that when he writes $Ht_i$ in Equation $(3)$, he really means $(kH)t_i$. Is this correct, or am I misinterpreting something?
Edit: For vector spaces, we have the relation $$ U\otimes (V\oplus W)\cong U\otimes V\oplus U\otimes W, $$ so it appears that the induced $G$-module should really be written $$ \text{ind}_H^G\cong \bigoplus_iU\otimes_H (kH)t_i. $$ So when Cohn writes an equality does he really mean an isomorphism?