I am searching for (two) presentations 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], in two separate cases:
REP. 1: $(ab)^4 = 1, (bc)^4 = 1, (ac)^2 = 1$;
REP. 2: $(ab)^3 = (bc)^3 = (ac)^3 = 1$.
Can anyone help me out here ? (Thanks !!)
This cannot be done. Let $G_1$ and $G_2$ be the groups $$G_1 = \langle a,b,c | a^2 = b^2 = c^2 = (ab)^4 = (bc)^4 = (ac)^2 \rangle$$ $$G_2 = \langle a,b,c | a^2 = b^2 = c^2 = (ab)^3 = (bc)^3 = (ac)^3 \rangle.$$
You are looking for surjections from the $G_j$ onto $PGL_3(\mathbb{F}_2)$, which is the simple group of order $168$.
Now, $G_1$ and $G_2$ are Coxeter groups and, in fact, they are two of the affine Coxeter groups -- types $\tilde{B}_2$ and $\tilde{A}_2$ respectively. They are known to have the alternate descriptions $D_6 \ltimes \mathbb{Z}^2$ and $D_8 \ltimes \mathbb{Z}^2$, where $D_{2m}$ is the dihedral group of order $2m$.
In particular, these two groups are solvable. However, $PGL_3(\mathbb{F}_2)$ is not solvable, so it cannot be a quotient of a solvable group.
There is a pretty direct argument here. Note first that $G = {\rm PGL}(3,2) = {\rm SL}(3,2)$ is a finite simple group of order 168. Suppose it were so generated.
Note next that $a$ inverts both $ab$ and $ac$, so normalizes $\langle ab, ac \rangle$. Now $bc = (ca)(ba)^{-1} \in \langle ab,ac \rangle$, so that $a$ normalizes $\langle ab,bc,ac \rangle.$ The reason I leave the redundant generator visible is that by symmetry, it follows that $b$ and $c$ both normalize $\langle ab,bc,ac \rangle.$ Thus $\langle ab,ac,bc \rangle$ is normal in $\langle a,b,c \rangle = G.$
Hence $\langle ab, bc, ac \rangle$ is either trivial or all of ${\rm PGL}(3,2)$, by simplicity.
Triviality of the subgroup $\langle ab,bc,ac \rangle$ yields $a = b = c$, so we have a contradiction.
Suppose then that $\langle ab,bc,ac \rangle = G.$ I give the complete argument in the case of REP2: let $x = ab, y = bc, z = ac.$ Then $x^{3} = y^{3} = z^{3}= 1$, by assumption, and $xy = z.$ The following argument was known to W. Burnside.
Now $\langle x^{-1}y, yx^{-1} \rangle$ is an Abelian normal subgroup of $G$, because $(xyx)(yxy) = 1$, as $xy$ has order $3$, yielding $yxy = (xyx)^{-1} = (x^{-1}y)(yx^{-1})$, while also $(yx^{-1})(x^{-1}y) = yxy$ because $x$ and $y$ have order 3.
Thus $x^{-1}y$ and $yx^{-1}$ commute. But $x(x^{-1}y)x^{-1} = yx^{-1},$ so that $x$ normalizes $\langle x^{-1}y, yx^{-1} \rangle$. Since $x^{-1}y$ certainly normalizes $\langle x^{-1}y, yx^{-1} \rangle,$ we see that $\langle x^{-1}y, yx^{-1} \rangle \lhd \langle x,y \rangle = G$, a contradiction.
There is a similar well-known argument to exclude the case of REP1, since then $\langle ab,ac,bc \rangle$ is a "$(2,4,4)$"-group
Later edit: Instead of using a generators and relations type argument, it is also possible to see via a character calculation that if $u,v,w$ are elements of order $4,4$ and $2$ respectively in $G$ with $uv = w$, then $\langle u,v \rangle$ is a cyclic $2$-group.
For $G$ has 6 complex irreducible characters of degrees $1,3,3,6,7,8$. Let $u$ be an element of order $4$ of $G$ and $w$ be an involution (there is a single class of elements of order $4$ and a single class of involutions in $G$). Only the irreducible characters $\chi$ of degrees $1,3,3$ and $7$ have $\chi(u)\chi(w) \neq 0$, and we find from the well-known character formula that the number of times $w$ is expressible as a product of two conjugates of $u$ in $G$ is $\frac{168}{16} [ 1 -\frac{1}{3}-\frac{1}{3} - \frac{1}{7}] = 2$. On the other hand, if we take $D = C_{G}(w)$ to be a Sylow $2$-subgroup in which $w$ is central, then $w$ is already expressible as $x^{2}$ and as $y^{2}$, where $x$ and $y$ are the two elements of order $4$ in $D$.
Hence whenever $w$ is expressible as a product of two elements of order $4$ in $G$, we see that these two elements of order $4$ are equal.
But rep 1 gives elements $ab, bc$ of order $4$ with $(ab)(bc) = ac$ of order $2$, and, as we have seen, we should have $\langle ab,bc \rangle = G$, a contradiction.
