# Let $G$ be a finite matrix group in $GL_2(Q)$ such that every matrix $A\in G$ has integer entries. Prove $A^{12}= I$ for each $A$.

Let $$G$$ be a finite matrix group in $$GL_2(Q)$$ (general linear group of $$2$$ by $$2$$ matrices with rational entries) such that every matrix $$A\in G$$ has integer entries. Prove that $$A^{12} = I$$ for every $$A \in G$$.

Attempt :

We have that $$A^k = I$$ for some natural $$k$$ since $$G$$ is finite. Then the minimal polynomial of $$A$$ divides $$x^{k} - 1$$. Also, the characteristic polynomial is of the form $$x^2 + ax + b$$ for some $$a,b$$. Thus, the minimal polynomial has either a root of the form $$x-c$$ where $$c$$ is an integer, or it is the characteristic polynomial itself. In the first case, we get $$A = I$$ or $$A =-I$$ since $$1,-1$$ are the only integer roots of unity. Hence $$A^2 = I$$.

In the second case, we have that either the roots of the characteristic polynomial are $$1,-1$$ in which case we get $$x^2 - 1$$, so $$A^2 = I$$ again. Otherwise, we have a complex root of unity and it's conjugate. This gives us $$b = 1$$ as it is the product of these roots, and $$a$$ is $$2 *$$ the real part. Hence we get $$A^2 + aA + I = 0$$ so $$A(A + aI) = -I$$. How do I use this to show $$A^{12} = I$$?

We also know that $$a$$ is an integer. Suppose that the root of unity $$\omega$$ is a root of $$x^2 + ax + b$$. Then we know that $$2 \operatorname{Re}(\omega) = a$$. But we have that $$-1 \leq \operatorname{Re}(\omega) \leq 1$$, and so we have that $$-2 \leq a \leq 2$$.

We know just consider all of the possible cases.

If $$a = \pm 2$$, then we have that $$\omega = \pm 1$$, which you have already dealt with.

If $$a = 0$$, then we have that $$\omega = i$$, which is a fourth root of unity. Note that we do not have that the minimal polynomial divides $$x^6 - 1$$, but it does divide $$x^{12} - 1$$.

If $$a = -1$$, then the characteristic polynomial is $$x^2 - x + 1$$, and $$\omega$$ is a sixth root of unity. (We have that $$x^2 - x + 1 \mid x^3 + 1 \mid x^6 - 1$$.)

If $$a = 1$$, then the characteristic polynomial is $$x^2 + x + 1$$, and $$\omega$$ is a third root of unity.

Notice above that if the minimal polynomial is $$x^2 + 1$$, then we do not have that $$A^6 = 1$$. This can in fact occur. Consider for example the matrix $$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$

Thus the problem as it currently stands is not true, but your original version before you edited it was correct. The exponent does need to be $$12$$.

• How do we know $a$ is an integer? Edit : Sorry, I'll change it back to 12. – Saad Oct 13 '18 at 10:30
• @Saad because $a=-\operatorname{Tr}(A)$ and $b=\det(A)$ – Hagen von Eitzen Oct 13 '18 at 10:34