I'm trying to understand the "easy" direction of the CFSG: namely, the proofs that the 18 infinite families and 26/27 sporadic groups are indeed simple. I'm working through Simple Groups of Lie Type by Carter, and The Finite Simple Groups by Wilson, which uses Iwasawa's Lemma to show the simplicity of the Mathieu groups, the Leech lattice groups, and $Fi_{22}$. However, I'm having trouble finding anything beyond that.
I'm happy to settle for definitions like "Let $J_1$ be the subgroup of $GL(20, \mathbb{F}_2)$ generated by [the matrices at https://brauer.maths.qmul.ac.uk/Atlas/v3/matrep/J1G1-f2r20B0]", if there's a way to use those representations to check the simplicity. I'm guessing there isn't really a way to do this for a general group other than applying the classification, but given that these groups are indeed simple, and lots is known about them, maybe there is some particular thing I could look at that would allow a computationally feasible verification of that fact.
For example, to show one of the groups G is perfect, no matter its size all I'd need to do is use generators $a, b$ with $a^2 = b^3 = (ab)^p$ with $p >= 7$ (which exist for the ones I care about according to https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-33/issue-3/On-Hurwitz-generation-and-genus-actions-of-sporadic-groups/10.1215/ijm/1255988653.full, and a lot of them are simply the standard generators listed in the ATLAS). Then I would define $G$ as the group generated by $a$ and $b$, provide explicit formulas* for $a$ and $b$ as products of commutators, and that would prove $[G, G] = G$.
Does anyone have any ideas for something like that that doesn't depend on the order of the group, but shows it's simple instead of perfect?
* \begin{align*} [b, a][a, bb]([a, b][bb, a][b, a][a, bb])^m &= babbaabbab(ababbbbabababbaabbab)^m \\ &= babab(abababababab)^m \\ &= b(ab)^{2+6m} = b(ab)^{pr} = b \text{ for some m and r} \end{align*}
\begin{align*} [a, b][bb, a]([b, a][a, bb][a, b][bb, a])^n &= ababbbbaba(babbaabbabababbbbaba)^n \\ &= abababa(babababababa)^n \\ &= a(ba)^{3+6n} = a(ba)^{ps} = a \text{ for some n and s} \end{align*}