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 $G$ 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 $\chi$. Let $(G, \mathcal{V}, \mathcal{S})$ be a regular topological map of type $(m,n)$ (where these are true orders). Recall that, when the map is regular we can think of $G$ 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 $G$ in what follows..

Write $V$ (resp. $E$, $F$) 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:

$\chi=V-E+F=|G|(\frac1{m}+\frac1{n}+\frac12) = -|G|\frac{mn-2m-2n}{2mn}.$

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 $m$ (resp. $2, n$).

Now write $[m,n]$ (resp. $(m,n)$) for the lcm (resp. gcd) of $m$ and $n$. Then we can rearrange the above equation to obtain that

$\frac{|G|}{[m,n]} = \frac{-2(m,n)\chi}{mn-2m-2n} = \frac{-2\chi}{[m,n]-2\frac{m+n}{(m,n)}}.$

The two equations displayed so far immediately yield some nice consequences. In what follows, if $k$ is an integer, $p$ a prime, write $|k|_p=p^s$ if $p^s$ is the highest power of $p$ that divides $k$.

Proposition 1: Let $p$ be a prime dividing the order of $G$. Either $p$ divides $\chi$ or a Sylow $p$-subgroup of $G$ is cyclic.

Proof: All of the statements that we make depend only on the equations given above. Let $P$ be a Sylow $p$-subgroup of $G$. Suppose that $p\nmid \chi$. Then either $p\mid[m,n]$ or $|G|_p=2$. But in the latter case $P$ is cyclic and we’re done. So we assume from here on that $p\mid [m,n]$.

Suppose that $|G|_p=|[m,n]|_p$. In this case $|G|_p=|m|_p$ or $|G|_p=|n|_p$; suppose, without lost of generality that $|G|_p=|m|_p$. Since we know that $G$ contains a cyclic group of order $m$ (a vertex stabilizer) we conclude that $P$ is cyclic as required.

Suppose that $|G|_p\neq |[m,n]|_p$; then $p\mid \frac{|G|}{[m,n]}$. Since $p\nmid\chi$ we conclude that $p=2$ and $\chi$ is odd. But this implies that $m$ and $n$ are odd and so $|G|_2=2$ and $P$ is cyclic as required.

QED

Now it is well-known in group theory that any group with a cyclic Sylow $2$-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 $\chi=2-2g$ where $g$ is the genus of the surface. (Incidentally I’d like a good explanation of that. I really don’t understand how $\chi$ and $g$ 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 $p$ be a prime dividing the order of $G$. Either $p^2$ divides $\chi$ or a Sylow $p$-subgroup of $G$ has a cyclic subgroup of index $p$.

Proof: Again all of the statements that we make depend only on the equations given above. Let $P$ be a Sylow $p$-subgroup of $G$. Suppose that $p^2\nmid \chi$; there are three possibilities: (a) $|G|_p=p$; or (b) $p=2$ and $|G|_2\leq 4$; or (c) $p\mid [m,n]$. The first two cases yield the result immediately, hence we assume that (c) holds.

Suppose that $|G|_p\leq p \cdot |[m,n]|_p$. In this case, WLOG, $|G|_p\leq p\cdot |m|_p$ or $|G|_p=|n|_p$. Since we know that $G$ contains a cyclic group of order $m$ (a vertex stabilizer) the result follows.

Finally suppose that $|G|_p > p \cdot |[m,n]|_p$. Then $p=2$ and $\chi=2a$ for some odd integer $a$. But now if either $m$ or $n$ are even, then $[m,n]-2\frac{m+n}{(m,n)}$ is even and we conclude that $|G|_2=2$ and the result follows. On the other hand if $m$ and $n$ are both odd, then $|G|_2=2$ and the result follows.

QED

I am most interested in applying Prop. 2 in the case $p=2$. In this situation we know explicitly what the group $G$ 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.