I am working through a question on an old character theory exam. I've answered the first two parts ok, but am now struggling on the third part. Here is the part that I can't do:

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I've computed the character table, and I get

$$\begin{array}{|c|c|c|c|c|c|} \hline & e & g_2 & g_3&g_4&g_5\\ \hline \chi_1 & 1 & 1& 1 & 1& 1\\ \hline \chi_2 & 1 & 1& -1 & -1& 1\\ \hline \chi_3 & 3 & -1& -1 & 1& 0\\ \hline \chi_4 & 3 & -1& 1 & -1& 0\\ \hline \chi_5 & 2 & 2& 0 & 0& -1\\ \hline \end{array}$$

so now I'm just confused about parts (ii) and (iii).

I know that every normal subgroup is an intersection of kernels of some of the irreducible characters, and that $\ker\chi_5=g_2^G$ (the second conjugacy class), but I'm not too sure where exactly to go from here. Any hints would be much appreciated!


1 Answer 1


From the character degrees you know that $|G|=24$. So there are $1$ or $4$ Sylow subgroups. Now $g_5$ has centralizer oforder $3$, so $g_5$ has order $3$, and $P=\langle g_5 \rangle$ is a Sylow $3$-subgroup of $G$. There must be $4$ Sylow 3-subgroups since otherwise the centralizer would be larger.

The conjugation action of $G$ of the set of the $4$ Sylow $3$-subgroups defines a homomorphism $\phi:G \to S_4$ with transitive image, and since no Sylow $3$-subgroup normalizes any other, the image of $\phi$ has order $12$ or $24$. But since $|N_G(P)|=6$ and $|C_G(P)|=3$, there is an element $g$ of order $2$ that normalizes but does not centralize $P$, so $\phi(g)$ is a transposition and hence ${\rm im}(\phi) = S_4$ so $G \cong S_4$.


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