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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.

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    $\begingroup$ You need to send $1$ to some number $x$ such that $x^8 = 1$. Have you considered some complex numbers? $\endgroup$
    – Joppy
    Commented Mar 31, 2022 at 2:42
  • $\begingroup$ @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. $\endgroup$ Commented Mar 31, 2022 at 2:44
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    $\begingroup$ 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. $\endgroup$ Commented Mar 31, 2022 at 2:51

1 Answer 1

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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.

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