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.
As you said, all irreducible complex representations of $\mathbb{Z}_8$ are one-dimensional.
Furthermore, if $f:\mathbb{Z}_8\to G$ is a homomorphism, then $f(\mathbb{Z}_8)$ is cyclic of order $d$, where $d$ divides $8$.
So look for cyclic subgroups of $\mathbb{C}^\times$ of order $1,2,4,8$
and map $\mathbb{Z}_8$ onto each.