Prove that any element of GL2(Z) of finite order has order 1,2,3,4, or 6.
Extend this to GL3(Z), and show that it fails in GL4(Z).
Thanks for your help!
Let X be an element of finite order in GL_2(Z) such that with .
Then the characteristic polynomial of X for eigenvalues is . Since is an eigenvalue of for a positive integer k, we see that for and .
Case 1. Both and are reals. In this case, for i=1,2 are +1 or -1. Thus it has order 1 or 2.
For order 1, .
For order 2, , , .
Case 2. and are complex numbers. Since they are roots of quadratic equations with real coefficients, they are conjugate to each other. Since they are complex roots of unity, .
If , then with or for k=1,2. Thus the order of X is 4 (verify this). If , then characteristic polynomial is . In this case, X can be . It has an order 3.
If , then characterist polynomial is . In this case, X can be . It has an order 6.
Thus any element of GL2(Z) of finite order has order 1,2,3,4, or 6.
the possible orders of elements in is again and however, the order of is equal to note that is the companion matrix of
the cyclotomic polynomial of order more examples are and the orders of
and are and respectively. also, and are the companion matrices of and respectively. finally, note that where is the
Euler totient function! extending this idea, we see that for any integer there exists with order