More math-blogging! Yay!
Today I have a proof I wrote recently, of a relatively fundamental fact: principal congruence subgroups of the modular group contain no elliptic elements.
The proof is available as a PDF or below the fold.
Theorem. The congruence subgroup 
contains elliptic elements if and only if 
, the modular group.
Proof. In three parts.
- It is easy to see that
contains elliptic elements. For instance, the following matrix is elliptic, with fixed points 
:
![\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]](http://memethief.com/wp-content/cache/tex_7981e0a1ad564ed12259d4f3e7cefd82.png)
- Assume that
The characteristic polynomial of
is 
. Thus, using the fact that 
, the discriminant is:
To show that
is elliptic, we aim to establish that its fixed points are two distinct complex conjugate points, ie, that its discriminant is negative. For this to be true it is necessary and sufficient for the trace 
of to be between -2 and 2. Since all variables here are integers, we make use of the stronger statement:
This implies that
or 
. The only possibilities are that 
and 
.
- Now we suppose that , and return to:
This implies that
, a contradiction.
Therefore, the only principal congruence subgroup that contains elliptic elements is the modular group itself. QED
The implications of this fact have yet to be fully realized. The question is still open as to whether it is possible to generate all the congruence subgroups of a given level, but the notion of decomposing subgroup generators into constituent generators, then manipulating those one by one is an interesting prospect.
