Let $G$ be the alternating group $A_5$, let $\chi_1$ be the trivial character of $G$, and let $\chi_2$ be the irreducible character of $G$ given by
where $\alpha = (\sqrt{5} + 1)/2$ and $\beta = (-\sqrt{5} + 1)/2$. The conjugacy class sizes, in order, are $1,20,15,12$, and $12$.
This question will use the character $\chi_2$ to construct the entire character table of $G$. Recall from the discussion about rotation angles in Assignment 8, Question 4 that if $x \in G$ is a 5-cycle, then $x$ is never conjugate to $x^2$. The characters $\chi_S$ and $\chi_A$ appearing in Proposition 19.14 of the book are known respectively as the symmetric square and the exterior square of the character $\chi$.
Show that the symmetric square, $\chi_{2, S}$, of $\chi_2$ is an integer-valued character of degree 6. Show that the trivial character $\chi_1$ is a constituent of $\chi_{2, S}$ (as defined in Definition 14.19). Show that $\chi_5:= \chi_{2, S}-\chi_1$ is an irreducible character of degree 5.
Here is my work so far:
By Proposition 19.13, $$\dim (\chi_{2, S}) = n(n+1)/2 = 3(3+1)/2 = 12/2 = 6.$$ To find the other values of the character, we use proposition 19.14, which says that $$\chi_{2,S}(g) = \frac{\chi^2(g) + \chi(g^2)}{2}.$$ Thus, we find that \begin{align*} \chi_{2, S}((1\:2\:3)) &= \frac{0^2 + 0^2}{2} = 0 \\ \chi_{2, S}((1\:2)(3\:4)) &= \frac{(-1)^2 + 3^2}{2} = 4 \\ \chi_{2, S}((1\:2\:3\:4\:5)) &= \frac{\alpha^2 + \beta^2}{2} = \frac{3/2 + \sqrt5/2 + 3/2 - \sqrt5/2}{2} = 3/2 \end{align*} which is NOT an integer! Where have I gone wrong?
You almost had the correct answer. In this case the permutation $g = (1 2 3 4 5)$ and $g^2 = (1 3 5 2 4)$. Thus, $\chi(g) = \alpha$ and $\chi(g^2) = \beta$. Now $$ \chi_{2,S}(g) = \frac{\chi^2(g) + \chi(g^2)}{2}= \frac{\alpha^2 + \beta}2 = \frac{(3+\sqrt{5})/2 + (1-\sqrt{5})/2}2 = 1.$$
Note the fact that $\chi(fg)\ne\chi(f)\chi(g)$ in general although $\rho(fg)=\rho(f)\rho(g)$ for the representation $\rho$ corresponding to $\chi$.