Let $G$ be a finitely generated free (nonabelian) group, $H$ a subgroup generated by some of the generators of $G$, and $a: G\to AG$ be the projection to the abelianization $AG:=G/[G,G]$.

Is it true that the normal closure $N(H)$ equal $a^{-1}(H)$?

In other words, can I rewrite every element in $H \cdot [G,G]$ as a product of elements of type $ghg^{-1}$?

Thanks for possible hint!

(Edit: I'm mostly interested in the case when $H$ is nontrivial, thx @Tsemo Aristide)


$N(H)$ normally denotes the normalizer of $H$, not its normal closure! I would denote its normal closure by $\langle H^G \rangle$.

The answer is usually no, because $H/\langle H^G \rangle$ is isomorphic to the free group on the generators of $G$ that are not in $H$, which is nonabelian provided that there are at least two such generators, but $a^{-1}(H)$ contains $[G,G]$, so $G/a^{-1}(H)$ is abelian.


It is not true, suppose $H=\{e\}$ the neutral element. $a^{-1}(1)=[G,G]$, but $H$ is normal, thus $H$ is equal to its normal closure.

  • $\begingroup$ Thanks, I'm stupid. However, if $H\neq \{1\}$, isnt' it less obvious? $\endgroup$ – Peter Franek Jul 24 '16 at 20:00
  • $\begingroup$ Isn't $G$ supposed to be a free group? $\endgroup$ – Max Jul 24 '16 at 20:44
  • 2
    $\begingroup$ You can't suppose that $G = [G,G]$ is not trivial, because it is not true! $\endgroup$ – Derek Holt Jul 24 '16 at 22:07

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