## If the first two lower central quotients of a group are cyclic, then the second derived subgroup is trivial

Show that if $G^\prime/G^{\prime\prime}$ and $G^{\prime\prime} / G^{\prime\prime\prime}$ are both cyclic then $G^{\prime\prime} = 1$.

We begin by proving a lemma.

Lemma: Let $G$ be a group and let $N \leq G$ be normal. Then $(G/N)^\prime = G^\prime N /N$. Proof: $(\subseteq)$ If $[xN,yN] \in G/N$ is a commutator, then $[xN,yN] = [x,y]N \in G^\prime N/N$. $(\supseteq)$ $G^\prime \leq G^\prime N$, so that $G/G^\prime N \cong (G/N)/(G^\prime N/N)$ is abelian. By Theorem 5.7 in the text, $(G/N)^\prime \leq G^\prime N/N$. $\square$

Now for the main result.

Note that $G^{\prime\prime\prime} \leq G$ is characteristic, hence normal, so that $G^{\prime\prime}/G^{\prime\prime\prime} \leq G/G^{\prime\prime\prime}$ is a cyclic, hence abelian, normal subgroup.

By this previous exercise, $(G/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}) \cong G/G^{\prime\prime}$ (via the Third Isomorphism Theorem) acts on $G^{\prime\prime}/G^{\prime\prime\prime}$ by conjugation on the left as follows: $(gG^{\prime\prime}) \cdot (a G^{\prime\prime\prime}) = (gag^{-1}) G^{\prime\prime\prime}$. Now conjugation is an automorphism of $G^{\prime\prime}/G^{\prime\prime\prime}$, and since this quotient is cyclic, its automorphism group is abelian. Thus, for all $g,h \in G$ and $a \in G^{\prime\prime}$, we have $(gh)a(gh)^{-1} G^{\prime\prime\prime} = (hg)a(hg)^{-1} G^{\prime\prime\prime}$. Via some arithmetic, we see that $[g,h] a [g,h]^{-1} a^{-1} \in G^{\prime\prime\prime}$; in particular, each element of $G^{\prime\prime}/G^{\prime\prime\prime}$ commutes with each generator of $G^{\prime}/G^{\prime\prime\prime}$, so that $G^{\prime\prime}/G^{\prime\prime\prime} \leq Z(G^{\prime}/G^{\prime\prime\prime})$.

Using the Third Isomorphism Theorem, $(G^\prime/G^{\prime\prime\prime})/Z(G^\prime/G^{\prime\prime\prime}) \cong ((G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))/(Z(G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))$ $\cong (G^\prime/G^{\prime\prime})/(Z(G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))$ is cyclic; hence $G^\prime/G^{\prime\prime\prime}$ is abelian, and using the lemma we have $1 = (G^\prime/G^{\prime\prime\prime})^\prime = G^{\prime\prime}/G^{\prime\prime\prime}$. Thus $G^{\prime\prime} = G^{\prime\prime\prime}$.