Some Sylow arguments
I want to outline some arguments concerning Sylow groups that are based on results in this paper of Conder, Potočnik and Širán. My arguments will look slightly different from those of CPS because I will restrict myself to orientable surfaces, and will define the group which is associated with my regular map according to the previous posts on this blog (the group in CPS is constructed differently – it is twice as large as ours).
For the purposes of this post all maps will be finite, hence (as we proved earlier) the corresponding surface is compact and we can define the Euler characteristic . Let be a regular topological map of type (where these are true orders). Recall that, when the map is regular we can think of as acting on the map via homeomorphisms of the surface (we can’t do this for non-regular maps), and this is how we will think of in what follows..
Write (resp. , ) for the number of vertices (resp. edges, faces) in our map. I assert that the following follows immediately from the work of Jones & Singerman that we have already discussed:
The first equality is the formula for the Euler characteristic, the second follows by regularity, the third by rearranging. Note that the second equality is based on the fact (which will be useful in its own right) that a vertex (resp. edge, resp. face) stabilizer is a cyclic subgroup of order (resp. ).
Now write (resp. ) for the lcm (resp. gcd) of and . Then we can rearrange the above equation to obtain that
The two equations displayed so far immediately yield some nice consequences. In what follows, if is an integer, a prime, write if is the highest power of that divides .
Proposition 1: Let be a prime dividing the order of . Either divides or a Sylow -subgroup of is cyclic.
Proof: All of the statements that we make depend only on the equations given above. Let be a Sylow -subgroup of . Suppose that . Then either or . But in the latter case is cyclic and we’re done. So we assume from here on that .
Suppose that . In this case or ; suppose, without lost of generality that . Since we know that contains a cyclic group of order (a vertex stabilizer) we conclude that is cyclic as required.
Suppose that ; then . Since we conclude that and is odd. But this implies that and are odd and so and is cyclic as required.
Now it is well-known in group theory that any group with a cyclic Sylow -subgroup is solvable. Thus Prop. 1 implies that any map on an orientable surface with odd Euler characteristic corresponds to a solvable group. Great!! Except that no orientable surfaces with odd Euler characteristic exist: in the orientable case where is the genus of the surface. (Incidentally I’d like a good explanation of that. I really don’t understand how and work in the world of topology.)
CPS have great success in using Prop. 1 to classify regular maps, with most of their attention focussed on non-orientable surfaces. But that story is for another day. For now let me do some bookwork and write down an extension of Prop. 1 that might come in handy to someone someday.
Proposition 2: Let be a prime dividing the order of . Either divides or a Sylow -subgroup of has a cyclic subgroup of index .
Proof: Again all of the statements that we make depend only on the equations given above. Let be a Sylow -subgroup of . Suppose that ; there are three possibilities: (a) ; or (b) and ; or (c) . The first two cases yield the result immediately, hence we assume that (c) holds.
Suppose that . In this case, WLOG, or . Since we know that contains a cyclic group of order (a vertex stabilizer) the result follows.
Finally suppose that . Then and for some odd integer . But now if either or are even, then is even and we conclude that and the result follows. On the other hand if and are both odd, then and the result follows.
I am most interested in applying Prop. 2 in the case . In this situation we know explicitly what the group looks like (see for instance this paper, and the papers it cites). I wanted to write a load more stuff on this but I’m out of time for now. More anon.