I am attempting to construct the character table for $\mathbb{Z}_8$. I know a few things off the bat:
Since $\mathbb{Z}_8$ is abelian, its conjugacy classes are singletons (i.e. we have eight classes)
Since $\mathbb{Z}_8$ is abelian, all irreducible representations are one-dimensional
So we note that for $\pi_1, \dots, \pi_8$ as a list of irreducoble representations, we always have $\pi_1(g) = 1$ is the trivial representation. We also know that $\pi_i(0) = 1$ since $\pi_i$ is a homomorphism. So we have so far
$$$$\begin{vmatrix} & \{0\} & \{1\} & \{2\} & \{3\} & \{4\} & \{5\} & \{6\} & \{7\} \\ \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \chi_2 & 1 & & & & & & & \\ \chi_3 & 1 & & & & & & & \\ \chi_4 & 1 & & & & & & & \\ \chi_5 & 1 & & & & & & & \\ \chi_6 & 1& & & & & & & \\ \chi_7 & 1 & & & & & & & \\ \chi_8 & 1 & & & & & & & \\ \end{vmatrix}$$ $$
Now, how can we fill in the remaining rows? I understand that, in other cases, we can utilize the fact that $\pi_i$ is a homomorphism and so, or example, with $\mathbb{Z}_2 \oplus \mathbb{Z}_2$, if we have $\pi(0,1) = -1$ and $\pi(1,0) = -1$, then $\pi(1,1) = \pi\big[ (1,0) + (0,1) \big] = \pi(1,0)\pi(0,1) = (-1)(-1)=1$.
So assume $\chi_2(1) = -1$ and then, working out the details, it follows that we have
$$\begin{vmatrix} & \{0\} & \{1\} & \{2\} & \{3\} & \{4\} & \{5\} & \{6\} & \{7\} \\ \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \chi_2 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \\ \chi_3 & 1 & & & & & & & \\ \chi_4 & 1 & & & & & & & \\ \chi_5 & 1 & & & & & & & \\ \chi_6 & 1& & & & & & & \\ \chi_7 & 1 & & & & & & & \\ \chi_8 & 1 & & & & & & & \\ \end{vmatrix}$$
But beyond this point, I am lost because:
If we assume $\chi_3(1) = 1$, then every value following it must be 1 (we already have this)
If we assume $\chi_3(1) = -1$, then every value following it must alternate (we already have this)
Does anyone have any advice regarding this issue? Thank you in advance.
