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}.

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  • adolfo  On May 13, 2012 at 11:08 pm

    G”=G”’ ¿cuándo supuso que G”’=1?

    • nbloomf  On August 20, 2012 at 2:06 pm

      This is an error in the text; Foote keeps an updated list of errata here.

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