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For a finite group with $|G|=p^{a}m$ and $p\nmid m$, a Sylow p-subgroup has order $p^a$. The Sylow theorems assert existence, containment of every p-subgroup in a Sylow subgroup, conjugacy of all Sylow subgroups, and $n_p\mid m$, $n_p\equiv 1(\mathrm{mod}p)$.

For Sylow subgroups of A6, $|A_6|=6!/2=360=2^3\cdot 3^2\cdot 5$. Explicit choices areTheir elements are even permutations, and their orders are the largest powers of five and three dividing $360$.

Every nonidentity element of $P_5$ is a five-cycle. There are $\left(\genfrac{}{}{0ex}{}{6}{5}\right)4!=144$ five-cycles, and distinct subgroups of order five meet only at the identity. Each contributes four such elements, givingFor $P_3$, each unordered partition of the six letters into two triples gives the subgroup generated by the two triple-cycles. It is independent of the chosen orientation of either cycle. There are $\frac{1}{2}\left(\genfrac{}{}{0ex}{}{6}{3}\right)=10$ such partitions. Different partitions give different subgroups, since the four single three-cycles in a subgroup identify the two triples. Every Sylow subgroup of order nine is conjugate to the displayed $P_3$, so every one arises from a partition in this way. HenceThe normalizer is $N_G(H)=\{g\in G:gH{g}^{-1}=H\}$. Its definition makes $H$ normal in $N_G(H)$. If $H$ is a Sylow subgroup of $G$, it is also a Sylow subgroup of $N_G(H)$, since $H\le N_G(H)\le G$ leaves no room for a larger power of $p$. The Sylow theorems make all Sylow subgroups of the normalizer conjugate there, and normality of $H$ makes every such conjugate equal to $H$. Thus $H$ is the unique Sylow subgroup of its normalizer.

Define $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. A group is a soluble group if this derived series reaches the identity after finitely many steps. Its derived length is the least $d$ with $G^{(d)}=1$.

Let $V=\{1,(12)(34),(13)(24),(14)(23)\}\le A_4$. This Klein four-group is a normal subgroup and Abelian; $A_4/V$ has order three, so $A_4^{\prime }\subseteq V$. The group $A_4$ is nonabelian, so its derived subgroup is nontrivial. Conjugation by a three-cycle permutes the three nonidentity elements of $V$ transitively. Since the derived subgroup is a normal subgroup, containing any one forces it to contain all three. Thus $A_4^{\prime }=V$, and $V^{\prime }=1$. Therefore $A_4$ has derived length two.

If $H\le G$, every commutator in $H$ is one in $G$, and induction gives $H^{(i)}\subseteq G^{(i)}$ for all $i$. Thus a subgroup of a solvable group is solvable, with no larger derived length.

The sign map gives $S_4^{\prime }\subseteq A_4$. The commutator $[(12),(23)]$ is a three-cycle. Normality of $S_4^{\prime }$ supplies every conjugate three-cycle, and three-cycles generate $A_4$. Hence $S_4^{\prime }=A_4$, so the complete derived series isEvery inclusion is strict, proving solvability and derived length three.