# 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.

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...).

• These calculations look tedious and thank you for it, your time specially 1729. ;) Oct 29 '12 at 15:50
• They are as non-tedious as you are going to get! Of course, you could just use GAP... Oct 29 '12 at 16:01
• Also, I skipped lots of steps when I was doing it. For example, I didn't write the presentation $\langle a, b, c, x, y;\ldots\rangle$, I got got rid of the $a$, $b$ and $c$ without thinking about it. Oct 29 '12 at 16:05
• I see that. Thanks again for your time. I am working on your calculations to know more and find my own for another time. :) Oct 29 '12 at 16:18
• @coffeemath: Maple can do groups?! I'm pretty sure all groups of order $p^2q$ are classified, but I cannot be sure (all groups of order $pq$ are, and this is a classic undergrad question). Also, all groups of order $p^aq^b$ are soluble (and so not simple if $ab\neq 0$). Oct 29 '12 at 19:59