Finding a generating set of the simple group of order 168

Question

I'm looking at the symmetries of some geometric object which I think should be the simple group $G$ of order 168. I have in my possession symmetries $\alpha$, $\beta$, and $\gamma$ of orders $2$, $3$, and $7$ respectively, which I think should be enough to generate all of $G$, but I am having a hard time verifying this.

Here are the relations I've computed among $\alpha$, $\beta$, and $\gamma$ so far:

• $\alpha\beta\alpha=\beta^{-1}$
• $\beta\gamma\beta^{-1}=\gamma^4$
• nothing recognizable in terms of $\alpha$, $\beta$, or $\gamma$ comes out of conjugating $\alpha$ by $\gamma$ or vice-versa.

The second relation shows that $\langle \beta,\gamma\rangle$ is a non-abelian group of order 21, and I've heard it mentioned that inside $G$, a subgroup of order 21 and an involution are enough to generate the whole group. Unfortunately, this mention did not come with a proof of this fact. What I know is that $G$ has a presentation $\langle x,y\mid x^2=y^3=(xy)^7=[x,y]^4=1\rangle$, so I'm hoping to be able to find $x$ and $y$ satisfying these relations from the elements I've described so far. I've been looking at setting $x=\alpha$ and then $y$ to be some conjugate of $\beta$, but I have yet to find any success. I'm looking for some help with this method of attack, or other strategies for showing that the group generated by my symmetries is actually $G$.

Answer 1

Your relations are compatible with a $PSL(3,2)$ (the simple group of order 168), as the following GAP computation shows:

gap> f:=FreeGroup("a","b","c");
<free group on the generators [ a, b, c ]>
gap> rels:=ParseRelators(f,"aba=B,bcB=c4,a2,b3,c7");
[ (a*b)^2, b*c*b^-1*c^-4, a^2, b^3, c^7 ]
gap> g:=f/rels;
<fp group on the generators [ a, b, c ]>
gap> q:=GQuotients(g,PSL(3,2));
[ [ a, b, c ] -> [ (4,6)(5,7), (2,4,6)(3,5,7), (1,2,4,3,6,7,5) ],
  [ a, b, c ] -> [ (2,4)(3,5), (2,4,6)(3,5,7), (1,2,4,3,6,7,5) ],
  [ a, b, c ] -> [ (2,6)(3,7), (2,4,6)(3,5,7), (1,2,4,3,6,7,5) ] ]

It takes a groups with the relations you give, and shows that there are three different ways that this would fit with the $PSL(3,2)$-quotient. Looking at the kernels of these quotients now would give you candidate relations to test for (in the hope that they will give enough restrictions to tie down the group).

If you want to try for the two-generators $x$ and $y$, you could try an explicit isomorphism, e.g. for the first quotient candidate. First, I ask to bring elements into normal form. This is not guaranteed to always work, but here it does:

gap> SetReducedMultiplication(g);

Now for the calculation:

gap> f2:=FreeGroup("x","y");
<free group on the generators [ x, y ]>
gap> psl:=f2/ParseRelators(f2,"x2,y3,(xy)7,[x,y]4");
<fp group on the generators [ x, y ]>
gap> Size(psl);
168

Since the command iso:=IsomorphismGroups(psl,Image(q[1])); does not yet work in GAP 4.11, we split it into two steps:

gap> isop:=IsomorphismPermGroup(psl);
[ x, y ] -> [ (1,2)(4,5), (2,3,4)(5,6,7) ]
gap> iso:=IsomorphismGroups(Image(isop),Image(q[1]));
[ (1,2)(4,5), (2,3,7,5)(4,6) ] -> [ (1,4)(3,6), (2,3)(4,6,5,7) ]
gap> iso:=isop*iso;
[ x, y ] -> [ (1,4)(3,6), (1,2,3)(4,5,7) ]
gap> List(GeneratorsOfGroup(psl),x->PreImagesRepresentative(q[1],
> ImagesRepresentative(iso,x)));
[ (a*c)^2, a*b*a ]

So in this case you could try $(\alpha\gamma)^2$ for $x$ and $\alpha\beta\alpha$ for $y$. (The other two quotients get a bit more complicated.)

Answer 2

The subgroup $H$ generated by elements of orders $2$, $3$ and $7$ has order divisible by $42$, so it has index $4$, $2$ or $1$ in $G$.

If it had index $4$ then there would be a nontrivial homomorphism $G \to S_4$, which contradicts the simplicity of $G$, and it is even clearer that it cannot have index $2$, so we must have $H=G$.