Say I am given a group with 4 conjugacy classes $C_1, C_2, C_3, C_4$ with orders 1,3,4,4 respectively. I will call the conjugacy class characters $\chi_1,\chi_2,\chi_3,\chi_4$ respectively.

I am given a character table with only information about $\chi_2$ Where $\xi^2 + \xi + 1 = 0$ as $\xi$ is a third root of unity.

\begin{array}{rrrrrrrrrrr} & C_1 & C_2 & C_3 & C_4 \\ \chi_0 & 1 & 1 & 1 & 1 \\ \chi_1 & & & & & & \\ \chi_2 & 1 & & \xi & & & \\ \chi_3 & & & & & & \\ \end{array}

Now I have made quite a mess trying to figure out all of the relations in order to figure out how to fill this character table in. The easiest is that the group has order 12 so I have to have the characters of degree 1,1,1, and 3 for $\chi_1,\chi_2,\chi_3,\chi_4$ respectively. Now I know about the row and column orthogonality - and I also believe that this table is for that of $A_4$ the alternating group.I also think that $\chi_3$ on $C_1$ has to be -1 as that conjugacy class has 3 elements and we would need:

$\chi_0 \cdot \chi_3: 1 \cdot 1 \cdot 3 + 3 \cdot 1 \cdot -1 + 4 \cdot 1 \cdot f + 4 \cdot 1 \cdot g = 3 - 3 + 4f + 4g = 4f + 4g = 0$

This of course would make $f = -g$. So of course $f = 1 \Rightarrow g = -1$ and we have the sum of the squares multiplied by the number of the elements of the conjugacy class equal to the order of the group, i.e. $3^2 + -1^2 + 1^2 + -1^2 = 12$ or possibly if $f = -1 \Rightarrow g = 1$, but I would have to figure out more of the elements of the rows and everything I come up with has so many variables I think that there has got to be a simpler way than what I am trying - I am missing some relationship(s) for the following:

\begin{array}{rrrrrrrrrrr} & C_1 & C_2 & C_3 & C_4 \\ \chi_0 & 1 & 1 & 1 & 1 \\ \chi_1 & 1 & a & b & c & & \\ \chi_2 & 1 & d & \xi & e & & \\ \chi_3 & 3 & -1 & f & g & & \\ \end{array}

I have found what I think is the solution here: Character Table

But there are things I don't understand, and I would like to know about as well as how to fill in the character table, according to the link:

  • "If the group G contains a normal subgroup H, any character of the quotient group G/H gives a character of G by composition with the canonical homomorphism G -> G/H. If the quotient happens to be Abelian, you can use the previous remark. In particular, the number of characters of degree 1 is the index of the derived group."

I don't see how to know, on a purely algebraic level and not knowing about the geometry if $G$ contains a normal subgroup $H$ I know if it does, than each for each $g \in G, gH=Hg \Rightarrow H = g^{-1}Hg$ hence normal subgroups ?. I don't know what this tells me about the character table, could someone give me an example using the table?

  • I know that the as the sum of the squares of the rows multiplied by the number of elements in the conjugacy class should equal the order of the group but when I look at the equation from the character table of $A_4$ I see for $\chi_2: 1 \cdot 1^2 + 3 \cdot 1^2 + 4 \cdot \xi^2 + 4 \cdot (\xi^{2})^2$, which as $\xi^2 = -1 - \xi$ hence $(-1 - \xi)(-1 - \xi) = 1 + 2 \xi + \xi^2$ I eventually end up with the full expression for the row of: $8 + 8 \xi + 8 \xi^2 = 8(1 + \xi + \xi^2) = 8(0) =0$.

But the order of the group is 12. So what happened? According to the link above: "The degrees of the irreducible characters divide the order of the group, and the sum of the squares of these degrees is equal to the order of the group"

From the information above, I can figure out the order of the derived subgroup $|[G,G]|$ and can describe it as a union of conjugacy classes. So far what I have: - If I let $x, y \in G$ then I get that $xy - yx = k$. I know that if $k = 0$ then we have an abelian group hence... there would be a single conjugacy class? But since there are 4 conjugacy classes $G$ is not abelian. Not sure where to go next on this one.

Is this even the right approach? What is the character table supposed to tell me about the derived subgroup?

Thank you for any help.

  • 1
    $\begingroup$ I would start with the observation that $\chi_2^2$ and $\overline{\chi_2}$ are both degree one irreducible characters. Thus they need to be equal (and also equal to the missing 1-dimensional character). The other observation I want to make is that because there are 3 1-dimensional characters the derived group $[G,G]$ must be of index 3. Because the derived group is a normal subgroup, it must consist of conjugacy classes $C_1$ and $C_2$. $\endgroup$ Commented Jun 28, 2015 at 19:31
  • 2
    $\begingroup$ Expanding on the comment of @Jyrki The quotient $G/[G,G]$ is isomorphic to the $1$-dimensional characters (under multiplication of characters), and as mentioned, any linear character must be $1$ on the conjugacy class corresponding to elements of even order. Further, multiplying a linear character with an irreducible character gives an irreducible character. So when there is only one irreducible character of a given degree, it must be $0$ on any conjugacy class where the linear characters are not all $1$. This fills out most of the table. $\endgroup$ Commented Jun 29, 2015 at 10:46
  • $\begingroup$ Thank you both (Jyrki and Tobias) greatly! A couple things. Now I know that if a subgroup $H$ of $G$ is normal if $\forall g \in G; gH = Hg$ and hence if two subgroups $H_1$ and $H_2$ are conjugate to eachother than $gH_1g^{-1} = H_2$ but don't see why $[G,G]$ must consist of $C_1$ and $C_2$. Next, I don't see why any linear character must be 1 on the conjugacy class corresponding to elements of even order. Lastly, while I see that there is only one irreducible character of order 3 and it is indeed 0 on $C_3$ and $C_4$, but why must it be 0? - Thanks much for your previous insights! $\endgroup$ Commented Jul 1, 2015 at 23:01

1 Answer 1


Please see Mathews and Walker textbook ( Mathematical Methods for Physics) for an elegant though dense argument on how to fill this.


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