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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 $P\cap G' \neq 1$.

Shaun
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Samantha Wyler
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    You could do it by induction on $|G|$. Since $G$ is nilpotent, $Z(G)\ne 1$, so by induction $G/Z(G)$ is cyclic, but this implies $G$ is abelian, and you can do that case. – Derek Holt Jul 22 '20 at 16:19
  • Doesn't it depend on how $Z$ and $[G,G]$ are related? – markvs Jul 22 '20 at 16:31
  • @DerekHolt the only problem is in order to use induction to show $G/Z(G)$ is cyclic I would need to know that $G/Z(G)\textbf{/}(G/Z(G)'$ is cyclic. – Samantha Wyler Jul 22 '20 at 16:50
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    Well it is because any quotient group of a cyclic group is cyclic. – Derek Holt Jul 22 '20 at 17:56

3 Answers3

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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.

Nicky Hekster
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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

Shaun
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markvs
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  • Doesn't the fact that $(G/G_{k + 1})/(G_k/G_{k + 1})$ is cyclic mean that $G_k/G_{k+ 1} \leq (G/G_{k +1 })'$, not necessarily that $G_k/G_{k+ 1} \leq Z(G/G_{k + 1})$? – Samantha Wyler Jul 22 '20 at 16:59
  • I use the fact that if the factor over the center is cyclic then the group is abelian. Or are you asking about a different part of the answer? – markvs Jul 22 '20 at 17:03
  • I'm trying to figure out why $G_k/G_{k+1}$ is central in $G/G_{k+1}$, but I think that might be a property of nilpotent groups now. – Samantha Wyler Jul 22 '20 at 17:06
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    That is by the definition of the lower central series. In fact $\gamma_{k+1}(G)$ is defined as the smallest subgroup with that property, – markvs Jul 22 '20 at 17:08
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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.

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.

David A. Craven
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  • True, I have edited my answer to reflect the fact that Derek's answer is even easier than my version. And the proof that maximal subgroups are normal is easy, by the way. Quotient out by the centre. Induction says it's normal. Thus the preimage is normal. It's either $M$, so $M<$ is normal, or $MZ(G)$, in whcih case $M$ is normal again as $MC_G(M)=G$. – David A. Craven Jul 23 '20 at 07:25