How to draw a Cayley Graph?

Question

I need to draw a Cayley graph, but I dont know how to do it. In particular i need the same graph as shown in the figure. Please help me.

I can do this for now:

\documentclass[tikz,border=5]{standalone}
\usetikzlibrary{lindenmayersystems}
\pgfdeclarelindenmayersystem{cayley}{
  \rule{F -> F [ R [F] [+F] [-F] ]}
  \symbol{R}{
    \pgflsystemstep=0.5\pgflsystemstep
  } 
}
\begin{document}
\begin{tikzpicture}
\draw l-system [l-system={cayley, axiom=[F] [+F] [-F] [++F], step=5cm, order=6}];
\end{tikzpicture}
\end{document}


\begin{document}
\begin{tikzpicture}
\draw l-system [l-system={cayley, axiom=[A] [+A] [-A] [++A], step=5cm, order=4}];
\end{tikzpicture}
\end{document}

Answer 1

One easy way to draw this is to do it recursively with TikZ!

For example like that:

\documentclass[tikz,border=5mm]{standalone}

\def\recdraw#1#2#3#4{
  %% #1 : Length of the current path
  %% #2 : Current position
  %% #3 : Current angle
  %% #4 : Number of recursion left
  \ifnum#4>0
  \begin{scope}[shift={(#2)}]
    \draw (0,0) -- (#3:#1);
    \recdrawB{.5*#1}{(#3:#1)}{#3}{\the\numexpr#4-1}
    \draw (0,0) -- (90+#3:#1);
    \recdrawB{.5*#1}{(#3+90:#1)}{\the\numexpr#3+90}{\the\numexpr#4-1}
    \draw (0,0) -- (-90+#3:#1);
    \recdrawB{.5*#1}{(#3-90:#1)}{\the\numexpr#3-90}{\the\numexpr#4-1}
  \end{scope}
  \fi
}

\def\recdrawB#1{
  \pgfmathsetmacro\foo{#1}\expandafter\recdraw\expandafter{\foo}
}


\begin{document}

\begin{tikzpicture}[scale=4]

  \foreach \r in {0,90,180,270}{
    \draw (0,0) -- (\r:1);
    \recdraw{.5}{\r:1}{\r}{7}
  }

\end{tikzpicture}

\end{document}

which gives

Note that it's not perfect, and the use of pic would certainly be prettier, but I haven't mastered pics yet... :(

By the way, the function \recdrawB is used here to do some calculation so we don't keep the ".5*.5*.5*..." (it speed up the process). But it would certainly be much faster (but a little less recursive) to find a way to avoid doing length computation (maybe with pics or with global variable).

The rest of the draw is not very hard, I'll do it later if you need (and if I have the time)!

Answer 2

This uses a local coordinate system that has the dimensions of the Cayley graph to annotate it. And then there is some mechanism that prints the group element depending on the location in the graph, such that one can use foreach loops. This mechanism is not necessarily good for going to higher order.

\documentclass[tikz,border=5]{standalone}
\usetikzlibrary{lindenmayersystems,calc}
\pgfdeclarelindenmayersystem{cayley}{% https://tex.stackexchange.com/a/223078
  \rule{F -> F [ R [F] [+F] [-F] ]}
  \symbol{R}{
    \pgflsystemstep=0.5\pgflsystemstep
  } 
}
\newcommand\mysymb[1]{\pgfmathtruncatemacro{\itest}{Mod(#1,8)}%
\ifcase\itest
\sigma%
\or
\tau%
\or
\sigma^{-1}%
\or
\tau^{-1}%
\or
\sigma^2%
\or
\tau^2%
\or
\sigma^{-2}%
\or
\tau^{-2}%
\fi}
\begin{document}
\begin{tikzpicture}
 \draw l-system 
 [l-system={cayley, axiom=[F] [+F] [-F] [++F], step=5cm, order=6}];
 \begin{scope}[shift={(current bounding box.center)},
  x={($(current bounding box.east)-(current bounding box.center)$)},
  y={($(current bounding box.north)-(current bounding box.center)$)}] 
  \path[nodes={fill=white}] (0,0) node[scale=2]{$e$}
  foreach \X in {0,...,3} {(\X*90:1/2) node[scale={2/sqrt(2)}] {$\mysymb{\X}$}
   foreach \Y [evaluate=\Y as \Ymod using {int(Mod(\Y,4))}] in {\the\numexpr\X-1\relax,\X,\the\numexpr\X+1\relax}
   {($(\X*90:1/2)+(\Ymod*90:1/4)$)node[scale={2/sqrt(3)}] 
   {$\ifnum\X=\Ymod 
   \mysymb{\the\numexpr\X+4}
   \else
   \mysymb{\X}\mysymb{\Ymod}
   \fi$}}};
  \draw[thick] (45:{sqrt(2)/5}) -- (45:{sqrt(1/2)*1.2}) -- (0,1.2)
   -- (-1.2,0) -- (0,-1.2)  -- (-45:{sqrt(1/2)*1.2}) -- (-45:{sqrt(2)/5}) --
   cycle; 
  \draw[thick] (135:{sqrt(2)/5}) -- (135:{sqrt(1/2)*1.1})
   -- (-1.1,0) --  (-135:{sqrt(1/2)*1.1}) --
   (-135:{sqrt(2)/5}) -- cycle;  
 \end{scope} 
\end{tikzpicture}
\end{document}