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To replicate ps aux (BSD style) in the AT&T version of ps, you have to use
ps -Ao user,pid,%cpu,%mem,vsz,rss,tt,stat,start,time,command
This command is compatible with scripts that expect the same output as ps aux.
The only difference is the sort order; ps aux sorts all processes by their start time whereas ps -Ao sorts them by PID.
ps -jef is a shorter command, but it will output different headers.:
As mentioned in another answer, this is unfortunately not true (another example is $GL_2(3)$ whose derived subgroup is $SL_2(3)$). One interesting thing to note is that if it was true, then this would give an easy proof of the conjecture that the Taketa inequality holds for all solvable groups.
Elaboration: The Taketa inequality is $\rm{dl}(G)\leq |\rm{cd}(G)|$ where $\rm{dl}(G)$ is the derived length of $G$ and $\rm{cd}(G) = \{\chi(1) | \chi\in \rm{Irr}(G)\}$ is the set of degrees of irreducible complex characters of $G$.
As mentioned, this inequality is conjectured to hold for all solvable groups, and the way it would follow if the derived subgroup was nilpotent is due to a theorem of Isaacs and Knutson, which states:
If $N$ is a normal nilpotent subgroup of $G$ then the derived length of $N$ is at most the number of degrees of irreducible complex characters of $G$ which do not have $N$ in their kernel.
Using the above theorem with $N = G'$ one immediately gets the conjecture.
A group is called supersolvable if it has a finite ascending series of normal subgroups such that each quotient group is cyclic. So all finitely generated nilpotent groups are supersolvable, but $S_3$ is supersolvable but not nilpotent; and all supersolvable groups are solvable, but $A_4$ and $S_4$ are solvable but not supersolvable.
It is true that $G$ supersolvable implies that $G'$ is nilpotent. The proof (in brief) is that, since cyclic groups have abelian automorphism groups, $G'$ centralizes all quotients in the normal series for $G$ with cyclic quotients, and hence the intersection of $G'$ with this series is a central series for $G'$.