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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!

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    $\begingroup$ The group is isomorphic to $SL(2,3)$ by the way. $\endgroup$
    – Nicky Hekster
    Aug 8, 2022 at 6:27
  • $\begingroup$ @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! $\endgroup$
    – Hetong Xu
    Sep 7, 2022 at 3:22
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    $\begingroup$ 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 $SL(2,3)$. $\endgroup$
    – Nicky Hekster
    Sep 7, 2022 at 9:50
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    $\begingroup$ 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. $\endgroup$
    – Nicky Hekster
    Sep 8, 2022 at 7:57
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    $\begingroup$ @NickyHekster Your comments are really helpful to me. Thank you so much! $\endgroup$
    – Hetong Xu
    Sep 8, 2022 at 10:22

1 Answer 1

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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.)

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  • $\begingroup$ Thank you! The thing is that I'm so unfamiliar with "semi-direct product". :( Thank you so much! $\endgroup$
    – Hetong Xu
    Aug 7, 2022 at 14:44
  • $\begingroup$ And thank you for pointing out the "Schur-Zassenhaus"! :) $\endgroup$
    – Hetong Xu
    Aug 7, 2022 at 14:45

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