I am searching for a presentation of the group $\mathbf{PGL}_3(\mathbb{F}_2)$ for which the generators are involutions $a, b, c$, and such that the following relations are present [among extra relations, of course]:
REP.: $(ab)^2 = 1, (bc)^3 = 1, (ac)^6 = 1$.
Can anyone help me out here (again) ? (Thanks !!)
$G = {\rm GL}(3,2)$ can't be generated this way. First, we may note that there is no element of order $6$ in $G$, so we must have one of : $ac = 1$ or $(ac)^2 = 1$ or $(ac)^3 = 1.$
If $ac = 1$ then $a = c$ and $G = \langle a,b \rangle $ is Abelian.
If $(ac)^2 = 1$, then $a$ and $c$ commute. But also $a$ and $b$ commute since $(ab)^2 = 1$. Then $a \in Z(G),$ contrary to the simplicity of $G$.
If $(ac)^3 = 1,$ then $G$ is a homomorphic image of $A_{4} \cong \langle x,y : x^{2} = y^3 = (xy)^3 = 1 \rangle$, since $G = \langle ab, bc , ac \rangle $ and we have $(ab)^2 = (bc)^3 = [(ab)(bc)]^3 = 1 $.
Recall from the previous question that $\langle ab,bc,ac \rangle$ is generated by any two of the listed generators, and is normalized by each of $a,b$ and $c$, so is normal in $G$.
