Nilpotent vs Solvability

Question

I know that if G is solvable, then all subgroups and factor groups of G are solvable.

I also know if N is normal in G, and N and G/N are solvable, then G is solvable.(which is kind of like the converse to the statement above)

But what about nilpotent? I know if G is nilpotent, then all subgroups and factor groups of G are nilpotent. This can be proved by induction.

But is the converse true? If not, what is a counterexample?

Answer 1

Consider the group $S_3{}{}{}{}{}{}$.

Answer 2

The easiest counterexample I can think of is the Borel subgroup $B$ of upper triangular matrices in ${\rm GL}_n(\mathbb R)$. It is solvable, but not nilpotent. Consider its unipotent radical $N$, the subgroup of upper triangular matrices with only 1 on the diagonal. It is nilpotent and normal in $B$. Then $B/N$ is isomorphic to subgroup of diagonal matrices $T$. Since $T$ is commutative, it is nilpotent. To summarize: We have an exact sequence $$ 1\longrightarrow N\longrightarrow B\longrightarrow T\longrightarrow 1 $$ with $N$ and $T$ nilpotent, where $B$ is not.

Answer 3

Maybe good to know : if $N \unlhd G$ and both $N$ and $G/N'$ are nilpotent, then $G$ is nilpotent (see for a proof here).