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 $3^3\cdot7|\xi(1)$; 



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> Question: Is $G$ non-solvable?
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Is $G$ non-solvable?
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 $25.$


