Let $G$ be a group and $H$ a subgroup and let $X_H$ be a representation of $H$. According to the book Algebra by Cohn the character of the representation of $G$ induced by $H$ is given by $$\tag{1} \chi_G(x)=\frac{1}{|H|}\sum_{y\in G}\dot{\chi}_H(yxy^{-1}) $$ where $$ \dot{\chi}_H(z):= \begin{cases} \chi_H(x), & \text{if } x\in H,\\ 0, & \text{otherwise}. \end{cases}. $$ Consider the alternating group $A_4$ with 4 conjugacy classes, containing elements $$ (1)\in K_1,\qquad (12)(34)\in K_2,\qquad (123)\in K_3,\qquad (132)\in K_4. $$ The cyclic group generated by $(123)$ is a subgroup, and its character table is given by
$$\begin{array}{c|ccc} \, & (1) & (123) & (132)\\ \hline \chi_1 & 1 & 1 & 1 \\ \chi_2 & 1 & e^{\frac{2\pi}{3}i} & e^{\frac{4\pi}{3}i} \\ \chi_3 & 1 & e^{\frac{4\pi}{3}i} & e^{\frac{2\pi}{3}i} \end{array}$$
Now I want to try and calculate $\chi_2^{A_4}(1)$ of the induced representation using Equation $(1)$, but I get $$ \chi_2^{A_4}(1)=\frac{1}{3}(\dot{\chi_2}(1)+3\dot{\chi_2}(1)+4\dot{\chi_2}(123)+4\dot{\chi_2}(132))=\frac{1}{3}(1+3+4e^{\frac{2\pi}{3}i}+4e^{\frac{4\pi}{3}i})=0 $$ (where each term in the second equality is multiplied by the size of the respective conjugacy class) This is not a useful character of $A_4$ so what am I doing wrong?
Edit: I think I might have found the issue. $\dot{\chi}_H(y(1)y^{-1})$ is equal to zero except for the following cases $$ \begin{aligned} &y=(1),\qquad y(1)y^{-1}=(1),\\ &y\in K_2,\qquad y(1)y^{-1}=(1),\\ &y=(123),\qquad y(1)y^{-1}=(123),\\ &y=(132),\qquad y(1)y^{-1}=(132). \end{aligned} $$
Using Eq. $(1)$, I then get $$ \chi_2^{A_4}(1)=\frac{1}{3}(\chi_H(1)+3\chi_H(1)+\chi_H(123)+\chi_H(132))=\frac{1}{3}(4+e^{\frac{2\pi}{3}i}+e^{\frac{4\pi}{3}i})=1, $$ which indicates that the induced representation is of degree 1.