Let $G$ be a finite group of order $2^7\cdot3^3\cdot5^2\cdot7$. Let $\mathrm{Irr}(G)$ be the set of all the irreducible $\mathbb{C}$-characters. Suppose that

(1) there is a character $\chi\in\mathrm{Irr}(G)$ such that $2^5\cdot7|\chi(1)$;

(2) there is a character $\theta\in\mathrm{Irr}(G)$ such that $5^2\cdot7|\theta(1)$;

(3) there is a character $\xi\in\mathrm{Irr}(G)$ such that $3^3\cdot7|\xi(1)$;

Question: Is $G$ non-solvable?

  • $\begingroup$ Doesn’t (3) contradict Frobenius? $\endgroup$ – Francois Ziegler Jul 6 '18 at 2:08
  • $\begingroup$ Sorry. It is my mistake. The order of $G$ is $2^7\cdot3^3\cdot5^2\cdot7$. $\endgroup$ – C. Simon Jul 6 '18 at 2:14
  • $\begingroup$ Is this for research, or is it an exercise you have been set? $\endgroup$ – Geoff Robinson Jul 6 '18 at 10:52

Such a group is not solvable. The existence of the irreducible characters given forces $F(G)$ to be a $2$-group (using Clifford's Theorem). If $G$ were solvable, that would imply that $G/O_{2}(G)$ is isomorphic to a subgroup of ${\rm GL}(n,2)$ for some $n \leq 7.$ But no such ${\rm GL}(n,2)$ has order divisible by $25.$

  • $\begingroup$ Thank you very much. I know you consider the action of $G/O_2(G)$ on $O_2(G)/\Phi(O_2(G))$. I want to know why the order of the kernel of this action is not divisible by $25$. $\endgroup$ – C. Simon Jul 7 '18 at 2:24
  • $\begingroup$ If we assume that $G$ is solvable, then once we know that $F(G) = O_{2}(G),$ we know that $C_{G}(O_{2}(G)) \leq O_{2}(G).$ We know further that no element of odd order can act trivially on $O_{2}(G)/\Phi(O_{2}(G)),$ so that the kernel of the action on that Frattini factor is exactly $O_{2}(G).$ This is all standard group theory ( Hall-Higman, etc). $\endgroup$ – Geoff Robinson Jul 7 '18 at 8:32
  • $\begingroup$ @GeoffRobinson If $K$ is the kernel of the action of $G$ on $V:=O_2(G)/\Phi(O_2(G))$, then $G/K$ acts faithfully on $V$ and $O_2(G) \leq K$. Then the Hall-Higman (en.wikipedia.org/wiki/Hall%E2%80%93Higman_theorem) says that Sylow $q$-subgroups of $G/K$ are non-abelian. So, the order of $G$ implies that $5^2.7$ divides $|K|$. How can we explain this contradiction? $\endgroup$ – user97635 Jul 23 '18 at 16:24
  • 1
    $\begingroup$ @user97635 : As I said before, you need to read the full statement of the Hall-Higman theorem more carefully. You also need to read about the Hall-Higman centralizer Lemma, which tells you (when $G$ is solvable with $O_{2'}(G) = 1$) not only that the subgroup $K$ above is contained in $O_{2}(G)$ ( hence is equal to $O_{2}(G)$). The contradiction, as given in the answer above, is that given the existence of irreducible characters as in the question, the assumption that $G$ is solvable leads to a contradiction. $\endgroup$ – Geoff Robinson Jul 23 '18 at 19:24

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