Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.2 Exercise 7.2.9
Let $\alpha = r + r^2 – 2s$ and $\beta = -3r^2 + rs$ be elements of the group ring $\mathbb{Z}[D_8]$. Compute the following: $\beta\alpha$, $\alpha^2$, $\alpha\beta – \beta\alpha$, and $\beta\alpha\beta$.
Solution: Evidently, we have $$\beta\alpha = -3 – 2r – 3r^3 + s + 6r^2s + r^3s$$ $$\alpha^2 = 5 + r^2 + 2r^3 – 4r^2s – 4r^3s$$ $$\alpha\beta – \beta\alpha = 2r – 2r^3 – s + r^2s$$ $$\beta\alpha\beta = 15r + 10r^2 + 7r^3 – 21s – 6rs – 5r^2s$$
