On the structure of a finite group of order $144$ Let $G$ be a finite group of order $144$. Suppose there is an irreducible $\mathbb{C}$-character $\theta$ such that $\theta(1)=9$.

QUESTION: Prove $G\cong A_4\times A_4$.

By using Magma, we know there is only one group of order $144$ with an irreducible $\mathbb{C}$-character $\theta$ of degree $9$. Now I want to prove this result without using Magma. 
 A: Since $G$ has an irreducible character of degree $9 = |G|_{3},$ we have $O_{3}(G) = 1,$ so $G$ has more than one Sylow $3$-subgroup. If $G$ has only $4$ Sylow $3$-subgroups, then $G$ has a normal subgroup of order divisible by $6$ with a normal Sylow  $3$-subgroup ( consider the permutation action of $G$ on $N_{G}(S)$ for $S \in {\rm Syl}_{3}(G)$ in the case $[G:N_{G}(S)] = 4.$ The image in $S_{4}$ has order dividing $24,$ so the kernel of the action  has order divisible by $6$, and (being a subgroup of $N_{G}(S)$) has a normal Sylow $3$-subgroup, which remains normal in $G$, contrary to $O_{3}(G) = 1).$ 
Hence $G$ does indeed have $16$ Sylow $3$-subgroups, and a self-normalizing Sylow $3$-subgroup. By Burnside's normal $p$-complement theorem, $G$ has normal $3$-complement of order $16.$ In particular, $G$ is solvable (we could also see this from Burnside's $p^{a}q^{b}$-theorem, but that seems like overkill). 
Let $Q = O_{2}(G),$ which is a Sylow $2$-subgroup of $G.$ Since $O_{3}(G) =1,$ we have $Q = F(G).$ Now $G/Q$ acts faithfully as a group of linear transformations on $Q/\Phi(Q),$ so that $G/Q$ is isomorphic to a subgroup of ${\rm GL}(m,2)$ for some $m \leq 4.$ Since $[G:Q]$ is divisible by $9,$ we must have $m =4,$ and $Q$ is elementary Abelian. Since $GL(4,2)$ contains no element of order $9,$ we see that $G/Q$ is elementary Abelian of order $9.$ Further, the action of a Sylow $3$-subgroup of $GL(4,2)$ on the natural module gives the desired structure for $G.$
