# Using character table to find normal subgroups

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: I've computed the character table, and I get

|           | $e$ | $g_2$ | $g_3$ | $g_4$ | $g_5$ |
|-----------|-----|-------|-------|-------|-------|
| $\chi_1$  | 1   | 1     | 1     | 1     | 1     |
| $\chi_2$  | 1   | 1     | -1    | -1    | 1     |
| $\chi_3$  | 3   | -1    | -1    | 1     | 0     |
| $\chi_4$  | 3   | -1    | 1     | -1    | 0     |
| $\chi_5$  | 2   | 2     | 0     | 0     | -1    |


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!

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 $G \cong S_4$.