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Let $G$ be a finite group and $x, y, z \in G$. A hyperbolic generating triple for $G$ is a triple $(x, y, z) \in G\times G\times G$ such that

• $\frac{1}{o(x)}+\frac{1}{o(y)}+\frac{1}{o(z)} <1$,
• $\langle x,y,z \rangle =G$, and
• $xyz=1$.

The type of a hyperbolic generating triple $(x, y, z)$ is the triple $(o(x), o(y), o(z))$.

My question is, how can I use GAP to determine these triples for a group and therefore their type? Take $PSL(2, 7)$ as an example.

I thought this might work, but I cannot seem to print the triples.

g:=Size(G);;
cl:=ConjugacyClasses(G);; n:=Size(cl);;
class:=[];; GenTrips:=[];;
for i in [1..n] do
  Add(class ,AsList(cl[i]));;
od;

for i in [1..n] do
  catch:=[];
  x:=class[i][1];;
  for j in [i..n] do
    for j2 in [1..Size(class[j])] do
      y:=class[j][j2];;
      z:=Inverse(x*y);;
      for k in [j..n] do
        if z in class[k] then
          trip:=[i,j,k];;
          if trip in GenTrips then else
          if g=Size(Group(x,y)) then
              Add(GenTrips ,trip);; Add(catch ,k);
            fi;
          fi; 
          break;
        fi;
        if Difference([j..n],catch)=[] then
          catch:=[]; break;
        fi;
      od;
    od;
  od;
od;

Thanks

Let $G$ be a finite group and $x, y, z \in G$. A hyperbolic generating triple for $G$ is a triple $(x, y, z) \in G\times G\times G$ such that

• $\frac{1}{o(x)}+\frac{1}{o(y)}+\frac{1}{o(z)} <1$,
• $\langle x,y,z \rangle =G$, and
• $xyz=1$.

The type of a hyperbolic generating triple $(x, y, z)$ is the triple $(o(x), o(y), o(z))$.

My question is, how can I use GAP to determine these triples for a group and therefore their type? Take $PSL(2, 7)$ as an example.

I thought this might work, but I cannot seem to print the triples.

g:=Size(G);;
cl:=ConjugacyClasses(G);; n:=Size(cl);;
class:=[];; GenTrips:=[];;
for i in [1..n] do
  Add(class ,AsList(cl[i]));;
od;

for i in [1..n] do
  catch:=[];
  x:=class[i][1];;
  for j in [i..n] do
    for j2 in [1..Size(class[j])] do
      y:=class[j][j2];;
      z:=Inverse(x*y);;
      for k in [j..n] do
        if z in class[k] then
          trip:=[i,j,k];;
          if not trip in GenTrips then
          if g=Size(Group(x,y)) then
              Add(GenTrips ,trip);; Add(catch ,k);
            fi;
          fi; 
          break;
        fi;
        if Difference([j..n],catch)=[] then
          catch:=[]; break;
        fi;
      od;
    od;
  od;
od;

Thanks

Let $G$ be a finite group and $x, y, z \in G$. A hyperbolic generating triple for $G$ is a triple $(x, y, z) \in G\times G\times G$ such that

• $\frac{1}{o(x)}+\frac{1}{o(y)}+\frac{1}{o(z)} <1$,
• $\langle x,y,z \rangle =G$, and
• $xyz=1$.

The type of a hyperbolic generating triple $(x, y, z)$ is the triple $(o(x), o(y), o(z))$.

My question is, how can I use GAP to determine these triples for a group and therefore their type? Take $PSL(2, 7)$ as an example.

I thought this might work, but I cannot seem to print the triples.

cl:=ConjugacyClasses(G);; n:=Size(cl);;
class:=[];; GenTrips:=[];;
for i in [1..n] do
Add(class ,AsList(cl[i]));;
od;

for i in [1..n] do
catch:=[];
x:=class[i][1];;
for j in [i..n] do
for j2 in [1..Size(class[j])] do
y:=class[j][j2];;
z:=Inverse(x*y);;
for k in [j..n] do
if z in class[k] then trip:=[i,j,k];;
if trip in GenTrips then else
if g=Size(Group(x,y)) then
Add(GenTrips ,trip);; Add(catch ,k);
fi;
fi; break;
fi;
if Difference([j..n],catch)=[] then
catch:=[]; break;
fi;
od;
od;
od;
od;

Thanks

Let $G$ be a finite group and $x, y, z \in G$. A hyperbolic generating triple for $G$ is a triple $(x, y, z) \in G\times G\times G$ such that

• $\frac{1}{o(x)}+\frac{1}{o(y)}+\frac{1}{o(z)} <1$,
• $\langle x,y,z \rangle =G$, and
• $xyz=1$.

The type of a hyperbolic generating triple $(x, y, z)$ is the triple $(o(x), o(y), o(z))$.

My question is, how can I use GAP to determine these triples for a group and therefore their type? Take $PSL(2, 7)$ as an example.

I thought this might work, but I cannot seem to print the triples.

g:=Size(G);;
cl:=ConjugacyClasses(G);; n:=Size(cl);;
class:=[];; GenTrips:=[];;
for i in [1..n] do
  Add(class ,AsList(cl[i]));;
od;

for i in [1..n] do
  catch:=[];
  x:=class[i][1];;
  for j in [i..n] do
    for j2 in [1..Size(class[j])] do
      y:=class[j][j2];;
      z:=Inverse(x*y);;
      for k in [j..n] do
        if z in class[k] then
          trip:=[i,j,k];;
          if trip in GenTrips then else
          if g=Size(Group(x,y)) then
              Add(GenTrips ,trip);; Add(catch ,k);
            fi;
          fi; 
          break;
        fi;
        if Difference([j..n],catch)=[] then
          catch:=[]; break;
        fi;
      od;
    od;
  od;
od;

Thanks

Let $G$ be a finite group and $x, y, z \in G$. A hyperbolic generating triple for $G$ is a triple $(x, y, z) \in G\times G\times G$ such that

• $\frac{1}{o(x)}+\frac{1}{o(y)}+\frac{1}{o(z)} <1$,
• $\langle x,y,z \rangle =G$, and
• $xyz=1$.

The type of a hyperbolic generating triple $(x, y, z)$ is the triple $(o(x), o(y), o(z))$.

My question is, how can I use GAP to determine these triples for a group and therefore their type? Take $PSL(2, 7)$ as an example.

Thanks

Let $G$ be a finite group and $x, y, z \in G$. A hyperbolic generating triple for $G$ is a triple $(x, y, z) \in G\times G\times G$ such that

• $\frac{1}{o(x)}+\frac{1}{o(y)}+\frac{1}{o(z)} <1$,
• $\langle x,y,z \rangle =G$, and
• $xyz=1$.

The type of a hyperbolic generating triple $(x, y, z)$ is the triple $(o(x), o(y), o(z))$.

My question is, how can I use GAP to determine these triples for a group and therefore their type? Take $PSL(2, 7)$ as an example.

I thought this might work, but I cannot seem to print the triples.

cl:=ConjugacyClasses(G);; n:=Size(cl);;
class:=[];; GenTrips:=[];;
for i in [1..n] do
Add(class ,AsList(cl[i]));;
od;

for i in [1..n] do
catch:=[];
x:=class[i][1];;
for j in [i..n] do
for j2 in [1..Size(class[j])] do
y:=class[j][j2];;
z:=Inverse(x*y);;
for k in [j..n] do
if z in class[k] then trip:=[i,j,k];;
if trip in GenTrips then else
if g=Size(Group(x,y)) then
Add(GenTrips ,trip);; Add(catch ,k);
fi;
fi; break;
fi;
if Difference([j..n],catch)=[] then
catch:=[]; break;
fi;
od;
od;
od;
od;

Thanks