# The character table of an abelian group

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.

• You need to send $1$ to some number $x$ such that $x^8 = 1$. Have you considered some complex numbers? Mar 31, 2022 at 2:42
• @Joppy I suppose I've not encountered character tables with complex characters, although I could not see why that wouldn't be possible, since every representation sends a group element to $\mathbb{C}^\times$. So the roots of unity are the key here, I would assume. Mar 31, 2022 at 2:44
• In that case consider filling in the character table of $\Bbb{Z}_4$ first. The fourth root of unity is more famous than the eighth. Mar 31, 2022 at 2:51

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.