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Hm, something that seems kind of cool is the coincidence $A_5 \cong SL_2(\mathbb{F}_{2^2})$ (the latter group is of order $(4^2 - 1)(4^2 - 4)/(4-1) = 60$). I haven’t looked at this carefully, but it must follow from the statement that there is a unique perfect group of order $60$.
Put in the correct theorem numbers from Lang’s Algebra. He proves that (except for the case $n=2$ and $F = \mathbb{Z}/(2)$ or $\mathbb{Z}/(3)$) the group $SL_n(F)$ is not only perfect but also that $PSL_n(F)$ is simple. For the case $PSL_2(\mathbb{F}_4) = SL_2(\mathbb{F}_4)/\pm I$, this does not mean there’s a simple group of order $30$, because in fact $1 = -1$ in $\mathbb{F}_4$!!
Thanks! I have made more formal pointer to the refernces. And copied over the statement of perfection of $SL_n$ over fields to special linear group.
Gave a simple argument for why $SL_2(\mathbb{F}_4)$ is “the” simple group of order $60$ (“the” in parentheses since there are nontrivial automorphisms); this and not $SL_2(\mathbb{F}_5)$ is the smallest example from the $SL$ class.
Another thing one can say is that finite cartesian products of perfect groups and infinite direct sums of perfect groups are perfect. (In both cases the argument is very simple.) But I don’t know what one can say for infinite cartesian products – I wasn’t able to google my way to an answer.
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