Determine the group structure from its character table: a group of order $24$ as an example

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!

• The group is isomorphic to $SL(2,3)$ by the way. Aug 8, 2022 at 6:27
• @NickyHekster Sorry for such a late reply, but may I ask how did you see that this is the group $\mathrm{SL}_2(\mathbb{F}_3)$? I searched information of the group $\mathrm{SL}_2(\mathbb{F}_3)$ on groupprops. Properties of $G$ I get from the character table coincides with the group $\mathrm{SL}_2(\mathbb{F}_3)$. So I believe that. But as character tables may not determine the group structure (e.g. $D_4$ and $Q_8$ case), how can I see mathematically (instead of by faith) that this group is indeed $\mathrm{SL}_2(\mathbb{F}_3)$? Thank you so much! Sep 7, 2022 at 3:22
• Good point! It requires some more inspection indeed. From the character table one can derive that $G$ is solvable of derived length 3. There is only one other group of order $24$ having this property (see groupprops.subwiki.org/wiki/Groups_of_order_24) : $S_4$. But that group has an integer valued character table. So your character table is unique in the sense that is characterizes $SL(2,3)$. Sep 7, 2022 at 9:50
• Not sure you are familiar with Itô's Theorem: if $A \unlhd G$ abelian, then $\chi(1) \mid |G:A|$ for all $\chi \in Irr(G)$. This shows that $G'$ cannot be abelian (the group has degree $2$ irreducible characters) and in fact $G''=Z(G)$. Hope this helps. Sep 8, 2022 at 7:57
• @NickyHekster Your comments are really helpful to me. Thank you so much! Sep 8, 2022 at 10:22

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).