Let $G$ be a finite nilpotent group and $G'$ its commutator subgroup. Show that if $G/G'$ is cyclic then $G$ is cyclic.

Question

So I thought the cleanest way to do this was to simply prove $G' = 1$ since if $G$ is cyclic $G' = 1$ and then $G \cong G/G'$, but I got no where with this.

My next idea was since $G$ is nilpotent I know it's the direct product of normal Sylow subgroups which commute with one another, so it suffices to show that each Sylow subgroup is cyclic. If $P$ is a Sylow $p$ subgroup then $PG'/G'$ is cyclic so by the second isomorphism theorem so is $P/P\cap G'$. So if $P$ intersects $G'$ trivial y then $P$ is cyclic. But I don't know what to do with the case where $P\cap G' \neq 1$.

Answer 1

Let $$G=\gamma_0(G)>[G,G]=\gamma_1(G)>...>\gamma_n(G)=1$$ be the lower central series of $G$. We know that $G/\gamma_1(G)$ is cyclic. Let $k>1$ be the maximal number such that $G/\gamma_k(G)$ is cyclic. If $k=n$, we are done. So suppose $k<n$. Then $G_k/G_{k+1}$ is central in $G/G_{k+1}$. Since $G/G_k\equiv (G/G_{k+1})/(G_k/G_{k+1})$ is cyclic, we have that $G/G_{k+1}$ is abelian, so $\gamma_1(G)\le \gamma_{k+1}(G)$, a contradiction. QED

Answer 2

Derek's answer in the comments is the fastest. Induction on $|G|$. Since $G$ is nilpotent $Z(G)>1$. The commutator subgroup of $G/Z(G)$ must contains (the image of) the commutator subgroup of $G$, whence $G/Z(G)$ satisfies the induction hypothesis. Thus $G/Z(G)$ is cyclic, whence $G$ is abelian. Thus $G$ is cyclic and $G$ is generated by one element.

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The 'correct' way to do this is to work with maximal subgroups. It suffices to show this for $p$-groups. Notice that $G$ is cyclic means it is $1$-generator, so let $x_1,\dots,x_n$ be a minimal generating set for $G$. Let $M$ be a maximal subgroup of $G$, necessarily normal and of index $p$. Then $G/M$ is cyclic of order $p$. But $G/G'$ is cyclic, so $M$ must be the unique maximal subgroup of $G$. By assumption, removing any single generator from the set results in a proper subgroup, whence is contained in $M$. But then this means that (unless $n=1$) all $x_i$ lie in $M$, a contradiction.

This generalizes into the Burnisde basis theorem, one of the foundational results of the theory of $p$-groups.

Answer 3

Let $G$ be finite and nilpotent. Then $G' \subseteq \Phi(G)$, the Frattini subgroup. So, if $G/G'$ is cyclic, then $G/\Phi(G)$ is cyclic. Hence, there is a $g \in G$ with $G/\Phi(G)=\langle \overline{g} \rangle$. But then $G=\langle g \rangle\Phi(G)=\langle g \rangle$, since the elements of the Frattini subgroup are non-generators.