$\DeclareMathOperator\PSL{PSL}$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 $\lvert N\rvert$?



Irreducible and faithful $\operatorname{PSL}_2(q)$-module
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.


