**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.2 Exercise 7.2.10**

Consider the following elements of the integral group ring $\mathbb{Z}[S_3]$: $$\alpha = 3(1\ 2) – 5(2\ 3) + 14(1\ 2\ 3)$$ and $$\beta = 6(1) + 2(2\ 3) – 7(1\ 3\ 2).$$ Compute the following elements: $\alpha + \beta$, $2\alpha – 3\beta$, $\alpha\beta$, $\beta\alpha$, and $\alpha^2$.

Solution: Evidently, $$\alpha + \beta = 6(1) + 3(1\ 2) – 3(2\ 3) + 14(1\ 2\ 3) – 7(1\ 3\ 2)$$ $$2\alpha – 3\beta = -18(1) + 6(1\ 2) – 16(2\ 3) + 28(1\ 2\ 3) + 21(1\ 3\ 2)$$ $$\alpha\beta = -108(1) + 81(1\ 2) – 21(1\ 3) – 30(2\ 3) + 90(1\ 2\ 3)$$ $$\beta\alpha = -108(1) + 18(1\ 2) + 63(1\ 3) – 51(2\ 3) + 84(1\ 2\ 3) + 6(1\ 3\ 2)$$ $$\alpha^2 = 34(1) – 70(1\ 2) – 28(1\ 3) + 42(2\ 3) – 15(1\ 2\ 3) + 181(1\ 3\ 2)$$