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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.

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    $\begingroup$ This is hardly a research problem, but I am not voting to close, because I don't see immediately how to do it. It would help if you provided some context - where does the question come from? Here are some thoughts. By considering the restriction of the character to normal subgroups, there can be no subgroup of index $2$ in $G$, Using that you can rule out the number of Sylow $3$-subgroups being $1$ or $4$, so there must be $16$, and hence $G$ has a normal $2$-complement $N$ of order $16$. That's as far as I have got - there is more than one group with that property ($5$ in fact). $\endgroup$
    – Derek Holt
    Commented Jun 23, 2018 at 10:55
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    $\begingroup$ It might be useful that $A_4 \times A_4$ is the only group of order $144$ with no normal subgroup of index $2$, a normal subgroup of order $16$, and trivial centre. $\endgroup$
    – Derek Holt
    Commented Jun 23, 2018 at 11:16
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    $\begingroup$ @Derek Holt: The irreducible character of degree 9 implies no Abelian subgroup of order greater than 16, so that is enough, given what you have done already. $\endgroup$
    – Geoff Robinson
    Commented Jun 23, 2018 at 11:35
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    $\begingroup$ I know how to prove this result without using Magma - using GAP instead :-) $\endgroup$
    – Dima Pasechnik
    Commented Jun 23, 2018 at 21:45
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    $\begingroup$ @DimaPasechnik Well it's nice to know that GAP and Magma are in agreement. about this! But you have to be careful because they are probably not genuinely independent computations. The library of small groups is common to both for example. $\endgroup$
    – Derek Holt
    Commented Jun 23, 2018 at 22:54

1 Answer 1

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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 $G.$

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