# Nilpotent groups are solvable

I know this should be obvious but somehow I can't seem to figure it out and it annoys me! My definition of nilpotent groups is the following: A group $G$ is nilpotent if every subgroup of $G$ is subnormal in $G$, or equivalently if $U<N_G(U)$ for all $U<G$. And my definition of solvable groups is that a group $G$ is solvable if $U'\neq U$ for all subgroups $1\neq U \le G$, where $U'$ is the commutator subgroup of $U$. My question is then: why are nilpotent groups solvable?

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You've marked this as "finite groups". Are you concerned solely with finite groups? Do you know that finite nilpotent groups are products of $p$-groups? – Arturo Magidin May 12 '12 at 23:45
@ArturoMagidin: The given definition for nilpotent works only for finite groups (same for solvable). – jug May 13 '12 at 16:16
@jug: Right; just like "product of $p$-groups" only works for finite groups. It should still be stated explicitly rather than implicitly, don't you think? – Arturo Magidin May 13 '12 at 22:49

Let $G$ be nilpotent and nontrivial. Since every maximal subgroup of $G$ must be normal, and if $M$ is maximal then $G/M$ has no proper subgroups, it follows that if $M$ is maximal then $G/M$ is a group of order $p$, hence abelian. Therefore, $[G,G]\subseteq M$, since $[G,G]$ is contained in any normal subgroup $N$ of $G$ such that $G/N$ is abelian. In particular, $[G,G]\neq G$. Now simply note that being nilpotent is inherited to subgroups, as proven below, to conclude that $[H,H]\neq H$ for all subgroups $H$ of $G$ when $G$ is nilpotent. Hence, if every subgroup of $G$ is subnormal ($G$ is nilpotent), then the commutator subgroup of $H$ is properly contained in $H$ for any nontrivial subgroup $H$ of $G$ ($G$ is solvable).

(If $H\leq G$ and $K$ is a subgroup of $H$, then $K$ is subnormal in $G$, so there exist subgroups $K\triangleleft K_1\triangleleft K_2\triangleleft\cdots\triangleleft K_m=G$. Intersecting the subnormal series with $H$ gives you a subnormal series fo $K$ in $H$, showing $K$ is subnormal in $H$ as well; thus, every subgroup of $H$ is subnormal, so subgroup of nilpotent is nilpotent).

Added. I am tacitly assuming above that $G$ has maximal subgroups; so it might fail for infinite groups in which every subgroup is subnormal. In the infinite case, the usual definition of "nilpotent" is via either the upper central series or the lower central series, and that of "solvable" via the derived series. In the case of the definition via the lower central series, proving solvability is very easy: recall that the lower central series of $G$ is defined inductively by letting $G_1=G$ and $G_{n+1}=[G_n,G]$; and a group $G$ is nilpotent if and only if $G_{n+1}=\{1\}$ for some $n\geq 1$. Now note that $G^{(2)}=[G,G]=G_2$, and if $G^{(k)}\subseteq G_n$, then $G^{(k+1)} = [G^{(k)},G^{(k)}] \subseteq [G_k,G]=G_{k+1}$. So if the lower central series terminates, then so does the derived series, proving that if $G$ is nilpotent then $G$ is solvable.

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 Dear Arturo, Sorry, I began my answer before yours was posted and then had to put it aside for a few minutes; now that I've posted it and taken a look at yours, I see that I followed identical lines to you. I guess it is the natural way to approach the question! Best wishes, – Matt E May 13 '12 at 0:17 @Matte: No worries; and you used the normalizer condition directly, whereas I didn't. I've added some words about proving it in the infinite/has no maximal subgroups case using the lower central series instead. – Arturo Magidin May 13 '12 at 0:40

To begin with, let $M$ be a maximal proper subgroup of $G$, which is nilpotent. Then (by your definition of nilpotence) we must have that $M$ is normal in $G$. Thus $G/M$ is a group with no proper subgroups, hence is cyclic of prime order, and so in particular is abelian. Thus $G$ has a non-trivial abelian quotient, and so $G'$ is a proper subgroup of $G$.
Suppose that $H$ is a subgroup of $G$, and that $U$ is a proper subgroup of $H$. By assumption $N_G(U)$ properly contains $U$. I claim that $N_H(U)$ (which equals $H \cap N_G(U)$) properly contains $U$ also. Indeed if not, i.e. if $N_H(U) = U$, then $U =H \cap N_G(U)$. Now consider $N_G(N_G(U))$: we see that $$H \cap N_G(N_G(U)) = N_H(H \cap N_G(U)) = N_H(U) = U.$$ Continuing, we find that $H \cap N_G(N_G( ... N_G(U) ...)) = U$. But since $N_G$ of any proper subgroup properly contains that subgroup, the iterated $N_G$s on the left hand side eventually reach $G$, and so we see that $H\cap G = U$. But $H \cap G = H$, and so $H = U$, contradicting our assumption that $U$ is a proper subgroup of $H$. Thus indeed $N_H(U)$ properly contains $U$, and so $H$ is again nilpotent.