$\DeclareMathOperator\PSL{PSL}$Let $G$ be finite group and $N$ be the unique minimal normal subgroup of $G$. Assume that $N$ is abelian and $G/N \cong \PSL(2,2^f)$. Is there any upper bound for $\lvert N\rvert$?
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1$\begingroup$ What is the irreducible and faithful module referenced in the title? $\endgroup$– LSpiceCommented Dec 9, 2020 at 19:18
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$\begingroup$ @LSpice, $N$, can be considered as an irreducible and faithful $PSl(2,2^f)$-module. $\endgroup$– SaraCommented Dec 9, 2020 at 20:16
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$\begingroup$ Ah, so module in the sense of admitting a group action, not a linear action. Why is it faithful? $\endgroup$– LSpiceCommented Dec 9, 2020 at 20:44
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3$\begingroup$ @LSpice: the group ${\rm PSL}(2,2^f)$ is simple, so if the action is not faithful, it is trivial, in which case either $N$ is simple or not a minimal normal subgroup. Also, in this case the question is answered by the theory of Schur multipliers. $\endgroup$– Alex B.Commented Dec 9, 2020 at 21:10
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The answer is "no", since for every sufficiently large prime $p$ there are simple non-trivial $\mathbb{F}_p[{\rm PSL}_2(\mathbb{F}_{2^f})]$-modules. You can take $N$ to be any such module and form the semi-direct product of $N$ and ${\rm PSL}_2(\mathbb{F}_{2^f})$. The following argument shows that actually all $N$ satisfying the hypotheses are of this form (although not all $G$ need be semi-direct products; there may be a non-split piece).
Firstly, $N$ must be a $p$-group for some prime $p$, since every $p$-Sylow of an abelian group is a characteristic subgroup, hence would be a proper subgroup that is normal in $G$. Secondly, if $N$ is a $p$-group, then its Frattini subgroup $\Phi(N)=[N,N]N^p$ is also characteristic, so must be trivial. This means that $N$ is an elementary abelian $p$-group, and therefore forms an irreducible $\mathbb{F}_p$-representation of ${\rm PSL}_2(\mathbb{F}_{2^f})$.
In particular, $|N|$ is unbounded, since you can take your favourite complex irreducible representation of ${\rm PSL}_2(\mathbb{F}_{2^f})$ and realise it in any "good" characteristic. However the dimension of $N$ as an $\mathbb{F}_p$-vector space is bounded.
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$\begingroup$ Doesn't this mean the answer is no because as p gets large enough the order of N goes to infinity $\endgroup$ Commented Dec 9, 2020 at 21:50
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$\begingroup$ @BenjaminSteinberg right, but only in the "obvious" way of just realising "the same" representations in different characteristics. The point is that you cannot have larger more complicated modules, e.g. ones of exponent $p^n$ for some $n\geq 2$. But yes, the answer to the literal question as asked is "no". $\endgroup$– Alex B.Commented Dec 10, 2020 at 0:21
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$\begingroup$ I think I agree with Benjamin here. The question was unambiguous, and the answer is no. You are guilty of the serious crime of adapting the question to make it more interesting (which of course would be a very bad idea if it was a question in an exam). $\endgroup$ Commented Dec 10, 2020 at 13:59
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$\begingroup$ @DerekHolt Ok, I have edited (I don't think I can ping Benjamin in the same comment, alas). $\endgroup$– Alex B.Commented Dec 10, 2020 at 15:04