Elliptic elements in congruence subgroups

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 $${}_{m}hatGamma$$ contains elliptic elements if and only if $${}_{m}hatGamma = Gamma$$, the modular group.

Proof. In three parts.

  1. It is easy to see that $$Gamma$$ contains elliptic elements. For instance, the following matrix is elliptic, with fixed points $$pm i$$:
    $$! left[begin{array}{cc} 0 & -1 \ 1 & 0 end{array}right] $$
  2. Assume that $$m>1$$. Let:
    $$! P = left[begin{array}{cc} a & b \ c & d end{array}right] = left[begin{array}{cc} alpha m + 1 & beta m \ gamma m & delta m + 1 end{array}right] in {}_{m}hatGamma quad ; quad alpha, beta, gamma, delta in mathbb{Z} $$

    The characteristic polynomial of $$P$$ is $$cz^2 + (d-a)z – b$$. Thus, using the fact that $$ad-bc = 1$$, the discriminant is:
    $$! (d-a)^2 + 4cb quad = quad (a + d)^2 – 4 $$

    To show that $$P$$ 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 $$(a+d)$$ of $$P$$ to be between -2 and 2. Since all variables here are integers, we make use of the stronger statement:
    $$! (alpha + delta) m in { -3, -2, -1 } $$

    This implies that $$m | 1, m | 2$$ or $$m | 3$$. The only possibilities are that $$m in {2,3}$$ and $$alpha + delta = -1$$.

  3. Now we suppose that $$m in {2,3}$$, and return to:
    $$! begin{align} ad – bc & = 1 \ (alpha m + 1)((- alpha – 1 ) m + 1) – beta gamma m^2 & = 1 \ (-alpha^2 – alpha – beta gamma) m^2 – m & = 0 \ (-alpha^2 – alpha – beta gamma) m & = 1 end{align} $$

    This implies that $$m | 1$$, 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.

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