Let $G$ be a finite group of order $240$.
If $G\cong C\_4\times A\_5$ or $C\_2\times C\_2\times A\_5$, then the all degrees of irreducible $\mathbb{C}$-characters of $G$ are
$
[1,1,1,1,~3,3,3,3,3,3,3,3, ~4,4,4,4,~5,5,5,5 ].
$


Conversely,
Suppose that $G$ is non-solvable, and the all degrees of irreducible $\mathbb{C}$-characters are
$
[1,1,1,1,~3,3,3,3,3,3,3,3, ~4,4,4,4,~5,5,5,5 ].
$



> 
> Question: $G\cong C\_4\times A\_5$ or $C\_2\times C\_2\times A\_5$?
> 
> 
> 



$G\cong C_4\times A_5$ or $C_2\times C_2\times A_5$?
$G$ is non-solvable, so must have $A\_{5}$ as a composition factor (as no other non-Abelian simple group has less than $168$). Hence $F(G)$ can have order at most $4$.


If $G$ has no component, then $F(G) = F^{\ast}(G)$ has order $4$, and is centralized by all elements of order $5$, contrary to $C\_{G}(F^{\ast}(G)) \leq F^{\ast}(G).$ 


Hence $G$ has a component (quasi-simple subnormal subgroup). This is isomorphic to either $A\_{5}$ or ${\rm SL}(2,5).$ If it is (isomorphic to) ${\rm SL}(2,5),$ it is then normal (it has index two), and ${\rm SL}(2,5)$ has an irreducible character of degree $6$. By Clifford's theorem, $G$ has an irreducible character of degree at least $6,$ which is precluded here.


Hence $G$ has a normal subgroup $L$ isomorphic to $A\_{5}.$ Each irreducible character of degree $3$ of $A\_{5}$ must extend to an irreducible character of degree $3$ of $G$ (since $G$ has no irreducible character of degree $6$ or more). Hence no element of order $5$ in $L$ can be conjugate to all its non-identity powers. Let $S$ be a Sylow $5$-subgroup of $L$. Then by a Frattini argument, $G = LN\_{G}(S) = LC\_{G}(S)$ (since $[N\_{G}(S):C\_{G}(S)] = [N\_{L}(S):C\_{L}(S)] = 2).$ Now $G$ induces inner automorphisms on 
$F^{\ast}(G) = L \times O\_{2}(G)$ so $G = F^{\ast}(G)$ has one of the listed structures.


