nilpotent group implies solvable group

Question

Can someone please do a simple proof of this: If a group, G, is nilpotent then it is solvable.

I'm pretty bad at math and am just trying to figure this out. Thank you very much!

Answer 1

Here is a nifty alternative proof. If $G$ is finite and nilpotent, it is the direct product of its Sylow subgroups, i.e. we can write $$G=P_1\times P_2 \times \cdots P_n$$ with $P_i$ a Sylow $p_i$-subgroup of $G$, for each prime divisor $p_i$ of $|G|$. Thus given any nonempty subset $\{i_1,\ldots , i_m\}$ of $\{1,\ldots, n\}$ we have that $P_{i_1}\times \cdots \times P_{i_m}$ is a subgroup of $G$. It follows by the converse to Hall's theorem that $G$ is solvable.

Answer 2

Finite case, use induction on the order of $G$: take $P \in Syl_p(G)$, then $P$ is normal, and, being a $p$-group, solvable. But $G/P$ is again nilpotent and hence solvable. Now both $P$ and $G/P$ are solvable and $G$ must be solvable.