Finding a new presentation for $H=\langle ab,abc\rangle$ Using Todd-Coxeter algorithm (a handy long calculations), I could find that the subgroup  $H=\langle ab,abc\rangle$ of group $$G=\langle a,b,c|abcab^{-1}=bcabc^{-1}=cabca^{-1}=1\rangle$$ is of index $1$ in $G$ and has a new presentation as $$H=\langle x,y|y^{-1}xy^2x^2=1,xy=yx^{-3}yx\rangle$$ May I ask here to have another simpler way, at least in this group. Thank you.
 A: Clearly $H=\langle ab, abc\rangle=\langle ab, c\rangle$ as $(ab^{-1})abc=c$. On the other hand, $b^{-1}=c(ab)c^{-1}$ by the first relation. Thus, $b, c\in H$. As $ab, b\in H$, $b\in H$ and so $G=\langle ab, abc\rangle=H$. We therefore have index one.
To get the presentation for $G$ in terms of $ab, abc$, we re-write the presentation we have for $G$ in terms of $ab$ and $abc$. Start by just doing this for $a$, $b$ and $c$,
$c=(ab)^{-1}(abc)=x^{-1}y$
$a=(ab)^{-1}(abc)(ab)(ab)^{-1}(abc)=x^{-1}y^2$
$b=(ab)^{-1}(abc)(ab)^{-1}(abc)^{-1}(ab)=x^{-1}yx^{-1}y^{-1}x$
Then rewriting $G$ in terms of $x$ and $y$ we get,
$H=\langle a, b, c, x, y; a=x^{-1}y^2, b=x^{-1}yx^{-1}y^{-1}x, c=x^{-1}y^2, abcab^{-1}=1, bcabc^{-1}=1, cabca^{-1}=1\rangle$
$\Rightarrow H=\langle x, y; x^{-1}y^2x^{-1}yx^{-1}y^{-1}xx^{-1}y^2x^{-1}y^2(x^{-1}yx^{-1}y^{-1}x)^{-1}=1, x^{-1}yx^{-1}y^{-1}xx^{-1}y^2x^{-1}y^2x^{-1}yx^{-1}y^{-1}x(x^{-1}y^2)^{-1}=1, x^{-1}y^2x^{-1}y^2x^{-1}yx^{-1}y^{-1}xx^{-1}y^2(x^{-1}y^2)^{-1}=1\rangle$
$\Rightarrow H=\langle x, y; x^{-1}y^2x^{-1}yx^{-1}yx^{-1}y^2x^{-1}yxy^{-1}x=1, x^{-1}yx^{-1}yx^{-1}y^2x^{-1}yx^{-1}y^{-1}xy^{-2}x=1, x^{-1}y^2x^{-1}y^2x^{-1}yx^{-1}y^{-1}x=1\rangle$
$\Rightarrow H=\langle x, y; yx^{-1}yx^{-1}yx^{-1}y^2x^{-1}yx=1, x^{-1}yx^{-1}y^2x^{-1}yx^{-1}y^{-1}xy^{-1}=1, yx^{-1}y^2x^{-1}yx^{-1}=1\rangle$
Tidy it up a bit and you should get the same as what you have (assuming that neither of us have made any mistakes in our working, which, for my part, isn't unlikely...).
