[Group Theory] Constructing Cayley Graph from Given Relations

[esorey]

Homework Statement

Show that there exists a group of order 21 having two generators $s$ and $t$ for which $s^3 = I$ and $sts^{-1} = t^2$. Do this exercise by constructing the graph of the group.

Homework Equations

Based on the given relations, we have $t^7 = I$.

The Attempt at a Solution

Since $s$ and $t$ have periods of 3 and 7, respectively, I know that the graph can be based on either 3 heptagons or 7 triangles. The back of the book has a solution based on 7 triangles, but I would like to construct a graph based on heptagons for some much-needed practice. I see that I need three concentric heptagons to give the 21 elements of the group. However, I am having a hard time understanding how to connect the vertices of the heptagons to satisfy $sts^{-1} = t^2$. I have had similar issues with the graphs of simpler groups which I solved by brute force. However, this group is complex enough that I do not want to do that. Is there some algorithmic way of seeing how the vertices must be connected by $s$? If not, how do I go about figuring out the proper configuration?

Thanks

 

[Office_Shredder]

You know that s has to connect a vertex of one heptagon to a vertex in a different heptagon. For sts^-1 = t² to hold, then going from one heptagon to the next, taking a step around in the t direction, then coming back to the first heptagon is the same as taking two steps around the original heptagon.

 

[esorey]

I figured it out! For some reason, I didn't realize that it didn't matter which two vertices you connect first, since from there you derive the rest. Thanks!