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I have the following character table. Note I assume that $\chi_i$'s are all irreducible.

$$ \begin{array}{|c|c|c|c|c|} \hline & C_1 & C_2 & C_3 & C_4 & C_5 \\ \hline \chi_0 & 1 &1 & 1& 1&1 \\ \hline \chi_1 & 1 & \zeta_3^2 & \zeta_3 & 1 & 1 \\ \hline \chi_2 & 1 & \zeta_3 & \zeta_3^2 & 1 & 1 \\ \hline \chi_3 & 3 & 0 & 0 & \zeta_7^3+\zeta_7^4+\zeta_7^6 & \zeta_7+\zeta_7^0+\zeta_7^5 \\ \hline \chi_4 & 3 & 0 & 0 & \zeta_7+\zeta_7^2+\zeta_7^4 & \zeta_7^{-1}+\zeta_7^{-2}+\zeta_7^{-4} \\ \hline \end{array} $$

I would like the show that there is a normal subgroup of order $7$.

Now I would I am looking at the last two columns and thought that for there to be normal subgroup of order $7$ they would all need to equal $1$ because surely,

$$\text{Normal subgroup}=\bigcap_{\chi_i} \{ g \in G: |\chi_i(g)|=1 \} $$

Is this a correct expression, if not, what is the correct way of finding normal subgroups?

Trajan
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  • Link to: http://math.stackexchange.com/questions/1153190/complete-the-character-table-of-group-of-order-21 – Trajan Feb 18 '15 at 16:50

1 Answers1

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The irreducible characters of an abelian group are of degree $1$, and the table shows that this one has two of degree $3$.

The sum of the squares of the integers in the first column is $21$, so that is the order of the group. Notice that $\chi_1$ is a rep of degree $1$: an element $g$ of the group is in its kernel iff $\chi_1(g)=1$. It follows that $\ker\chi_1$ is the union of $C_1$, $C_4$ and $C_5$. This is a normal subgroup. Can you find its order?

Mariano Suárez-Álvarez
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  • Yeah.......15?? – Trajan Feb 18 '15 at 16:55
  • Where did you get that number from? As the group has order 21, a subgroup most certainly cannot have order 15, you see. – Mariano Suárez-Álvarez Feb 18 '15 at 16:56
  • I can see that was a silly response now – Trajan Feb 18 '15 at 16:57
  • a) this is a non-abelian group so not sure what you mean by the first sentence. b) why are you considering just $\chi_1$? – Trajan Feb 18 '15 at 17:12
  • He is considering $\chi_1$ because the kernel of $\chi_1$ is the normal subgroup you are looking for. – Derek Holt Feb 18 '15 at 17:32
  • As the order of a subgroup has to divide the order of the group wouldn't we have that the order of the kernel of the group homomorphism (which is a subgroup) would have to divide the order of the group (Lagrange's Theorem). So as the order of the group is 21, we would have that the order of the kernel has to be 3 or 7. I am not sure how to know if it is 3 or 7 though. – Relative0 Sep 19 '15 at 18:16
  • @MarianoSuárez-Alvarez ; If I understand correctly, the kernel of any representation $\chi$ is the union of all of its conjugacy classes in which $\chi (g) = 1$. Now with $\chi_1$ as it is a linear character (degree 1) we have the three conjugacy classes which are also of degree 1 in $\chi_1$'s kernel. How does one find its order though? Just adding together the number of elements in each of the conjugacy classes? (Since I don't know the number of elements in each it might be obviously impossible by Lagranges theorem), but how would one find the order of $\chi_1$? Thanks. – Relative0 Sep 19 '15 at 18:39