Quotient of Coxeter group Since the group $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10} \rangle$ seems to have resisted attacks of some powerful programs, I will turn to a group that seems to be a bit easier to analyse. The group
$$H := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^3, (bc)^7, (abc)^{19} \rangle$$
only has one of the simple groups that are a quotient of $G$, $J_1$. Therefore it seems it will be easier to analyse. 
Adding the relation $(abcbc)^i$ gives the trivial group for all $i$ less than 25 except for 15, where it gives $J_1$. What is the group
$$I := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^3, (bc)^7, (abc)^{19}, (abcbc)^{25} \rangle?$$ 
I have checked on magma using knuth bendix, and it takes too long, so magma stops the calculation.
 A: As you may know, in the family of groups
$H^{m,n,p} := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^m, (bc)^n, (abc)^p \rangle$
considered by Coxeter, there is only one remaining case for which finiteness has not been decided, $H^{3,7,19}$, which is your group $H$. It is possible that $H \cong J_1 \times {\rm PSL}(2,113)$, since only two finite simple quotients are known. The kernel of the homomorphism onto $J_1$ is perfect. I haven't quite proved it, but I think the kernel onto ${\rm PSL}(2,113)$ is also perfect.
A homomorphism onto ${\rm PSL}(2,113)$ is induced by the map onto ${\rm SL}(2,113)$
$$a \mapsto \left(\begin{array}{rr}58& 57\\ 50& 55\end{array}\right),
b \mapsto \left(\begin{array}{rr}55& 69\\ 2& 58\end{array}\right),
c \mapsto \left(\begin{array}{rr}100& 67\\ 43& 13\end{array}\right).$$
The image of $abcbc$ under this map has order $57$.
With the ACE coset enumerator, I have been able to show that adding the extra relation $(abcbc)^n$ gives the trivial group for $n=25,26,27,28,29$, but I've got stuck on $n=30$.
