Let $G$ be a group. Put $D_0=G$ and, for $i\geq 1$, put $D_i=[D_{i-1}, D_{i-1}]$ (the commutator subgroup). Show that if there exists $r\geq 0$ such that $D_r={1}$ then $G$ is solvable.
I know that given such a $D_r$ I need to construct a finite series of subgroups each normal in the next. I assume this will look like $$D_0 \unlhd D_1 \unlhd \dots \unlhd D_r \unlhd G $$
And I know that each quotient $D_{i+1}/D_i$ will be abelian, but how will I know that each of the subgroups is normal in the next?
Thanks!
You want to show $[G,G]$ is always normal in $G$. This subgroup is generated by commutators, so it suffices you show that conjugating a commutator gives a product of commutators. Now,
$$caba^{-1}b^{-1}c^{-1} = (caba^{-1} c^{-1}b^{-1})(bcb^{-1}c^{-1})$$
is a product of commutators, that is
$$c[a,b]c^{-1} = [ca,b][b,c]$$
Alternatively, note that
$$caba^{-1}b^{-1}c^{-1}= (cac^{-1})(cbc^{-1})(ca^{-1}c^{-1})(cb^{-1}c^{-1})$$
that is, $$c[a,b]c^{-1} = [cac^{-1},cbc^{-1}]$$
and more generally, as noted in the comments,
$$\phi[a,b] = [\phi(a),\phi(b)]$$ for any morphism of groups $\phi$. This states the commutator subgroup is more than normal, it is completely characteristic, that is, stable under any group endomorphism.