Quotient group of the free group on the alphabet of a string.

Question

For instance, let $\Sigma = \{a,b\}$, and consider $F = F(\Sigma)$, the free group.

Let $s$ be a string, say $s = ab$ and consider the subgroup $H \leqslant F$, $H = \langle t: t \nleqslant s\rangle$ where $\leqslant$ on strings indicates "is a substring of". Then is $H$ normal?

Let $h \in H$, ie $h \nleqslant s$, and let $f \in F$. Then $fhf^{-1}$ is certainly in $H$ as you can't form a substring via concatenation if it involves any non substring, so $fhf^{-1} \in \{t \nleqslant s\}$ is in the generating set of $H$ so is in $H$. Therefore $F/H$ is a group.

What does $F/H$ look like!? It must be $\{\epsilon, a,b, ab\} \approx \Bbb{Z}_4$ right?

Does this group have a name in literature?

Answer 1

Your subgroup $H$ is all of $F$. For instance, if $s=ab$, then $aba$ and $abab$ are both in $H$, and hence so is $b=(aba)^{-1}(abab)$. Similarly, $a\in H$ as well. You can easily construct similar examples for any value of $s$.

Answer 2

If you pick $s = ab$ as your string, $h = b^{-1}ab$, as your not substring and $f=b$ as your other string, then $$fhf^{-1} = bb^{-1}abb^{-1} = a \leq ab = s.$$