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If $G$ is a connected solvable Lie-group, then $[G,G]$ is nilpotent. The corresponding statement for Lie algebras follows from Lie's theorem, and it then follows from connected Lie groups by exponentiation.

Is the statement also true for finite groups? I can't find a counterexample, but I didn't try that hard.

Motivation: I'm prepping to teach a group theory and Galois theory course, so I'm brainstorming interesting questions about finite groups. Whenever I come up with one I can't answer, I plan to toss it up here or at MO.

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4 Answers 4

up vote 14 down vote accepted

The symmetric group $S_4$ is solvable, but $[S_4, S_4] = A_4$ is not nilpotent.

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An interesting observation: finite and countable groups with discrete topology are exactly zero-dimensional Lie groups, but they are not connected and their Lie algebra is trivial 8) –  Alexei Averchenko Jan 8 '13 at 16:07

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.

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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 $G'$.

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I just want to throw in that there is actually a term for groups with nilpotent derived subgroups; they are called nilpotent-by-abelian. An equivalent condition is that $G'\leqslant \text{Fit}(G)$, which isn't always true in a solvable group.

In particular all solvable groups with $\ell_F(G)>2$ are counterexamples, which is another easy way to see that the property does not hold for $S_4$ (which is a $2$-Frobenius group, and thus has Fitting length $3$).

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