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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?

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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?

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  • $\begingroup$ Yeah.......15?? $\endgroup$ – Permian Feb 18 '15 at 16:55
  • $\begingroup$ Where did you get that number from? As the group has order 21, a subgroup most certainly cannot have order 15, you see. $\endgroup$ – Mariano Suárez-Álvarez Feb 18 '15 at 16:56
  • $\begingroup$ I can see that was a silly response now $\endgroup$ – Permian Feb 18 '15 at 16:57
  • $\begingroup$ a) this is a non-abelian group so not sure what you mean by the first sentence. b) why are you considering just $\chi_1$? $\endgroup$ – Permian Feb 18 '15 at 17:12
  • $\begingroup$ He is considering $\chi_1$ because the kernel of $\chi_1$ is the normal subgroup you are looking for. $\endgroup$ – Derek Holt Feb 18 '15 at 17:32

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