Finding a group with minimal generators and then a subgroup generated by these as an index two normal subgroup. Given a group with seven generators and seven relations, each of length 3, how can I use GAP to find the group generated by only three of its generators?
For example,
$$G = \langle a,b,c,d,e,f,g \mid abc, abd, acd, bef, cge, dfg, egf\rangle.$$
Furthermore, How can I then find an index two normal subgroup $H$ of this and the relations of $H$?
I'm trying to explore if some particular groups that are presented in "Star Graphs, Projective Planes and Free Subgroups in Small Cancellation Groups" by Edjvet & Howie can have a mixed Beauville structure.
I have this so far for the group above.
F:=FreeGroup(7);; a:=F.1;; b:=F.2;; c:=F.3;; d:=F.4;; e:=F.5;; f:=F.6;; g:=F.7;; 
rels := [abc, abd, acd, bef, cge, dfg, egf];; 
G := F/rels;; 
S:=SimplifiedFpGroup(G);
epi:=EpimorphismPGroup(S,2,3);;
img:=Image(epi);;
G23:=Image(epi); 
Size(G23); 
IdGroup(G23);
List(PCentralSeries(img),Size);

The idea is to take a group G and find a maximal 2-quotient of class 3 of that group, say S.  Then similarly, I need a 2 indexed normal subgroup of S, let's call it H, that is is also a maximal 2-quotient group of class 3.
I need to find both S and also the presentation of H to do this.
Any ideas?
 A: Using GAP here is overkill - we can do it by hand.
Firstly, we can use Tietze transformations here to find a presentation of $G$ with only two of the given generators (and $2<3$). I mentally worked out that this could be done easily because any of the first three relations can be used to remove $a$, while $bef$ can be used for $b$, $cge$ for $c$, $dfg$ for $d$ and $egf$ for $e$. However, in the following working I spotted a few more obvious moves which I did instead.
\begin{align*}
G &= \langle a,b,c,d,e,f,g \mid abc, abd, acd, bef, cge, dfg, egf\rangle\\
&\cong \langle a,b,c,d,e,f,g \mid a=c^{-1}b^{-1}, abd, acd, bef, cge, dfg, egf\rangle\\
&\cong \langle b,c,d,e,f,g \mid c^{-1}b^{-1}bd, c^{-1}b^{-1}cd, bef, cge, dfg, egf\rangle&\text{remove $a$ generator}\\
&\cong \langle b,c,d,e,f,g \mid c^{-1}d, c^{-1}b^{-1}cd, bef, cge, dfg, egf\rangle&\text{tidy up, spot that $c=d$}\\
&\cong \langle b,d,e,f,g \mid d^{-1}b^{-1}dd, bef, dge, dfg, egf\rangle&\text{remove $c$ generator}\\
&\cong \langle b,d,e,f,g \mid b^{-1}d, bef, dge, dfg, egf\rangle&\text{tidy up, spot that $b=d$}\\
&\cong \langle d,e,f,g \mid def, dge, dfg, egf\rangle&\text{remove $b$ generator}\\
&\cong \langle d,e,f,g \mid d=f^{-1}e^{-1}, dge, dfg, egf\rangle&\text{rearrange}\\
&\cong \langle e,f,g \mid f^{-1}e^{-1}ge, f^{-1}e^{-1}fg, egf\rangle&\text{remove $d$ generator}\\
&\cong \langle e,f,g \mid f^{-1}e^{-1}ge, e=fgf^{-1}, egf\rangle&\text{rearrange}\\
&\cong \langle f,g \mid f^{-1}(fgf^{-1})^{-1}g(fgf^{-1}), fgf^{-1}gf\rangle&\text{remove $e$ generator}\\
&\cong \langle f,g \mid g^{-1}f^{-1}gfgf^{-1}, fgf^{-1}gf\rangle&\text{tidy up}
\end{align*}

However, $G$ has no index-two subgroup. To see this, recall that such a subgroup is normal so we can consider the quotient group $G/H$, which is cyclic of order two. As $G/H$ is abelian, the map from $G\twoheadrightarrow G/H$ factors through the maximal abelian quotient of $G$, called the abelianisation $G^{ab}$ of $G$, so factors through the map $G\twoheadrightarrow G^{ab}$. So lets first abelianise $G$:
\begin{align*}
G^{ab}&\cong \langle f,g \mid g^{-1}f^{-1}gfgf^{-1}, fgf^{-1}gf, [g, f]\rangle\\
&\cong \langle f,g \mid gf^{-1}, g^2f, [g, f]\rangle\\
&\cong \langle g \mid g^3\rangle&\text{remove $f$ generator}\\
\end{align*}
Hence, every abelian quotient of $G$ is a quotient of the cyclic group of order $3$. Hence, $G$ does not have an index $2$ subgroup (but it does have an index $3$ subgroup).
