Here, the difficulty escalates. You must prove that a subset of an R-module is a submodule (closed under addition and scalar multiplication). Quotient modules are introduced, requiring you to verify that the module operations on cosets are well-defined.
Solution: It is clear that $\sim$ is reflexive and symmetric. To prove transitivity, let $x, y, z \in X$ such that $x \sim y$ and $y \sim z$. Then there exist $g, h \in G$ such that $y = gx$ and $z = hy$. Therefore, $z = h(gx) = (hg)x$, and we have $x \sim z$. dummit and foote solutions chapter 10
. While the concepts may seem familiar at first, the solutions in this chapter reveal the deep complexities that arise when the underlying scalars come from a ring instead of a field. The Core Themes of Chapter 10 Here, the difficulty escalates