I tried using sagemath. But I don't think sagemath is supporting character table of multiplicative groups of $(Z/nZ)^\times$. Also it would be great if you can suggest a way to calculate character table of $(Z/9Z)^\times$ which I can generalise into character table of $(Z/p^2Z)^\times$, where p is prime.

Thanks in advance.

  • 1
    $\begingroup$ All the $(\Bbb Z/p^2\Bbb Z)^\times$ are cyclic groups. $\endgroup$ – Angina Seng Aug 24 '18 at 19:01
  • $\begingroup$ All I need to find out is whether $(Z/p^2Z)^\times$ have an irreducible representation with character values {1,-1}. It is true when p=2. But I am not able to figure out when p>2. $\endgroup$ – user586874 Aug 24 '18 at 19:08
  • $\begingroup$ Yes: the group is cyclic of even order: you can use the Legendre symbol. $\endgroup$ – Angina Seng Aug 24 '18 at 19:09
  • $\begingroup$ I read about Legendre symbol just now but didn’t get the relationship between this and characters. Could you please elaborate ,or give me some references? $\endgroup$ – user586874 Aug 24 '18 at 19:26
  • $\begingroup$ The group's abelian, so all its irreps are $1$-dimensional, i.e., maps to $\mathbb{Z}_n$ for some $n$. $\endgroup$ – anomaly Aug 24 '18 at 19:29

I'll break this into steps using spoilers so you can try for yourself.

First, calculate what $G=(Z/9Z)^\times$ is.

To do this use the fact that the multiplicative subgroup has order $\phi(9)=6$ where $\phi$ is Euler's totient function, by doing some basic calculations we find the group has the elements $\{1,2,4,5,7,8\}$ and is isomorphic to a cyclic group of order 6.

Now we know what the group is we can apply basic character theory:

We know the group is abelian, so it must have $|G|=6$ conjugacy classes, and thus 6 irreducible representations, as our group is Abelian every representation is linear and thus a character. We immediately know one of these is the trivial representation and by the representation theory of abelian groups, the other representations are $1$-dimensional and the values take sixth roots of unity.

We can obtain these through lifting from the normal subgroups

For example, we can lift two nontrivial characters from the quotient group of $G/\{1,8\}$ and a non-trivial character from the quotient $G/\{1,4,7\}$.

The full table:

$\begin{array}{c|ccccc} &1&8&4&7&2&5\\ \hline ρ_1&1&1&1&1&1&1\\ ρ_2&1&-1&1&1&-1&-1\\ ρ_3&1&1&\zeta_3&\zeta_3&\zeta_3&\zeta_3^2\\ ρ_4&1&-1&\zeta_3^2&\zeta_3&\zeta_6^5&\zeta_6\\ ρ_5&1&1&\zeta_3&\zeta_3^2&\zeta_3^2&\zeta_3\\ ρ_6&1&-1&\zeta_3&\zeta_3^2&\zeta_6&\zeta_6^5\\\end{array}$

where $\zeta_k=e^{\frac{2i\pi}{k}}$.

  • $\begingroup$ Oh , got it! Thanks! $\endgroup$ – user586874 Aug 24 '18 at 19:22

Since you asked about doing it in Sagemath, note that a similar answer to this is available at this question.

sage: H = Integers(9)
sage: G = H.unit_group()
sage: G
Multiplicative Abelian group isomorphic to C6
sage: GG = gap(G)
sage: C = gap.CharacterTable(GG)
sage: gap.Display(C)

     2  1   1  1  1  1   1
     3  1   1  1  1  1   1

       1a  6a 3a 2a 3b  6b

X.1     1   1  1  1  1   1
X.2     1  -1  1 -1  1  -1
X.3     1   A /A  1  A  /A
X.4     1  -A /A -1  A -/A
X.5     1  /A  A  1 /A   A
X.6     1 -/A  A -1 /A  -A

A = E(3)^2
  = (-1-Sqrt(-3))/2 = -1-b3

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