Let $G$ be a finite group and the following is its character table (of irreducible $\mathbb{C}$-representations): $$ \begin{matrix} &g_1=1&g_2&g_3&g_4&g_5&g_6&g_7\\ \hline \chi_1 &1&1&1&1&1&1&1\\ \chi_2 &1&1&1&\omega^2&\omega&\omega^2&\omega \\ \chi_3 &1&1&1&\omega&\omega^2&\omega&\omega^2 \\ \chi_4 &2&-2&0&-1&-1&1&1\\ \chi_5 &2&-2&0&-\omega^2&-\omega&\omega^2&\omega\\ \chi_6 &2&-2&0&-\omega&-\omega^2&\omega&\omega^2\\ \chi_7 &3&3&-1&0&0&0&0\\ \end{matrix}$$
My question is how to prove: $G$ is the semi-direct product of its Sylow 2-subgroup and its Sylow 3-subgroup.
My knowledge on this group:
- The order of $G$: $24$. [By the square sum of the first column]
- Number of elements $m_i$ in each conjugacy class $\mathcal{C}_{g_i}$ with representative $g_i$: $(m_i)=(1,1,6,4,4,4,4)$.
- Kernel of each irreducible repn $\pi_i$ with character $\chi_i$: $$ \ker \pi_1=G, \, \ker \pi_2=\ker \pi_3 = \mathcal{C}_{g_1} \cup \mathcal{C}_{g_2} \cup \mathcal{C}_{g_3}, \, \ker \pi_7 = \mathcal{C}_{g_1} \cup \mathcal{C}_{g_2} $$ and the remaining representations are faithful (i.e. with trivial kernel).
- Normal subgroups: $\{1\}, \mathcal{C}_{g_1} \cup \mathcal{C}_{g_2}, \mathcal{C}_{g_1} \cup \mathcal{C}_{g_2} \cup \mathcal{C}_{g_3}, G$. They are of order $1,2,8,24$ respectively.
- Commutator subgroup: $[G,G] = \mathcal{C}_{g_1} \cup \mathcal{C}_{g_2} \cup \mathcal{C}_{g_3}$ of order $8$.
- Center: $Z(G)= \mathcal{C}_{g_1} \cup \mathcal{C}_{g_2}$ of order $2$.
- Sylow $2$-subgroup (of order $8$): there is already a normal subgroup $[G,G]$ of order $8$. So this is the unique Sylow $2$-subgroup of $G$. Call it $P$.
- Sylow $3$-subgroups (of order $3$): since there is no normal subgroup of order $3$, there are more than one Sylow $3$-subgroup. By Sylow theorem, there are four Sylow $3$-subgroups, which are all isomorphic to $C_3$, the cyclic group of order $3$. Call them $Q_1, Q_2, Q_3, Q_4$.
BUT I got stuck here to go any further to the show $G = P \rtimes Q_i$ for some $i$.
Thank you all for your help!
As you know that $P$ is normal, all that is left to prove that $Q_1 P = G$ and $Q_1 \cap P =1$, using the characterization of internal semidirect products. Both of these follow from order considerations (i.e. Lagrange).
More generally, if a Sylow subgroup in a group is normal, we get a semidirect product decomopsition by Schur-Zassenhaus. (Though that's overkill here.)