1
$\begingroup$

I am trying to understand the Cayley graph for the group $D_3$, which from Mathematica, I got: enter image description here

I tried to get the multiplication table in Mathematica:

$\left( \begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 1 & 4 & 3 & 6 & 5 \\ 3 & 5 & 1 & 6 & 2 & 4 \\ 4 & 6 & 2 & 5 & 1 & 3 \\ 5 & 3 & 6 & 1 & 4 & 2 \\ 6 & 4 & 5 & 2 & 3 & 1 \\ \end{array} \right)$

where the first row and column are the table headings.

How does this multiplication table relate to the graph?

Thanks!

$\endgroup$
2
+100
$\begingroup$

The table and graph are related as follows. Let $X=\{1,2,3,4,5,6\}$ and let $S=\{2,4\}$. The table is missing its "headings" so that the actual table looks like this $$ \begin{array}{c|cccccc} & 1 & 2& 3& 4& 5&6\\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 2 & 1 & 4 & 3 & 6 & 5 \\ 3 & 3 & 5 & 1 & 6 & 2 & 4 \\ 4 & 4 & 6 & 2 & 5 & 1 & 3 \\ 5 & 5 & 3 & 6 & 1 & 4 & 2 \\ 6 & 6 & 4 & 5 & 2 & 3 & 1 \\ \end{array} $$ where $i\cdot j$ is equal to the number in the $i$th row and $j$th column (so for example $4\cdot 2 =6$ and $2\cdot 4=3$). Next notice that $S$ generates $X$ under the above multiplication. To see this note that $2^2=1,2,4,4^2=5,2\cdot 4=3,2\cdot 4^2=6$ are all the elements of $X$. Now let us turn to the graph, the vertices are the elements of $X$ and there is a red edge from $x$ to $y$ if $x\cdot 2 =y$. Similarly there is purple edge from $x$ to $y$ if $x\cdot 4=y$.

$\endgroup$
  • $\begingroup$ Hi. Thank you for that. Could you please explain, though, why you set $S = \{2,4\}$? $\endgroup$ – Thomas Moore Sep 27 '16 at 13:26
  • $\begingroup$ You get a different Caley graph for each generating set. Since there are only two colours we see that for the above graph the generating set has two elements. To find out what the generating set is, you can look at the edges coming out of the identity element which in this case is $1$. $\endgroup$ – Nex Sep 27 '16 at 13:42
  • $\begingroup$ If you do this you will see that the graph above is obtained using the generating set I called $S$. $\endgroup$ – Nex Sep 27 '16 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.