My lecture book Algebra Vol. 2 by Cohn defines the induced $G$-module in the following way:
Let $H$ be a subgroup of $G$ an consider a right $H$-module $U$. From it we can form a right $G$-module $$\tag{1} \text{ind}_H^GU=U^G =U\bigotimes_HkG $$ called the induced $G$-module
He then writes the following:
The subspace $U\otimes H=\{u\otimes 1|u\in U\}$ of $U^G$ is an $H$-module in a natural way: for any $h\in H$ we have $(u\otimes 1)h=uh\otimes 1$. More generally we can define the subspace $U\otimes Ha=\{u\otimes a|u\in U\}$ of $U^G$ as an $H^a$-module, where $H^a=a^{-1}Ha$, by the rule $$\tag{3} (u\otimes a)x=uaxa^{-1}\otimes a\quad \text{for }x\in H^a. $$
Question: I assume that $U\otimes H$ is really $U\otimes kH$ (where $kH$ is the group algebra), which consists of elements of the form $u\otimes h$. I think the reason why we can write it as $U\otimes H=\{u\otimes 1|u\in U\}$ is that $$\tag{2} u\otimes h=u\otimes h\cdot 1=uh\otimes 1=u'\otimes 1, $$ is that true? Also, the reason why we have $(u\otimes 1)h=uh\otimes 1$ is that because we consider $kH$ strictly a left module in this case?