# elegant way to identify the order of a group's element - $A \in (GL(2,\mathbb{R}), \; \cdot \;)$

I'm looking for an elegant way to identify and justify in writing the order of a group's element.

$$A \in (GL(2,\mathbb{R}), \; \cdot \;) \quad , \quad A= \left( \begin{array}{cc} 0&1\\ -1&1\\ \end{array} \right)$$

$$ord \; A := min\{ A^k = E_2 \; | \; k \in \mathbb{N} \backslash \{0\} \}$$

I just started calculating $A$, $A^2$, $A^3$, ... and compared the result to $E_2$. So far so good, the result is $ord \; A = 6$.

But do I need to calculate and write down all prior powers to show that $A^k \neq E_2$ for all $k < 6$?

Lagrange doesn't help here either because $ord \; (GL(2,\mathbb{R}), \; \cdot \;) = \infty$ and therefore every $k \in \mathbb{N}\backslash \{0\}$ is divisor of $\infty$.

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You could just calculate the eigenvalues of your matrix. If your matrix is similar to a diagonal matrix (which it is if it has finite order) the order is just the lcm of all orders of eigenvalues. But since this is homework you should probably prove this first. Edit: Eigenvalues in the algebraic closure that is. –  Sebastian Schoennenbeck Feb 4 '12 at 10:18
Another way would be to find the smallest $n$ for which the minimal polynomial of your matrix divides $x^n-1$. This can also be done by calculating the eigenvalues in the algebraic closure. –  Sebastian Schoennenbeck Feb 4 '12 at 10:20
It is enough to calculate $A^2$, $A^3$ and $(A^3)^2 = A^6$. Then since $A^6 = 1$, the order of $A$ divides $6$. Because $A^2$ and $A^3$ are not the identity, we see that the order of $A$ is $6$. –  Mikko Korhonen Feb 4 '12 at 11:44
Why is it that the order of $A$ divides $6$? Lagrange says that the order of a subgroup divides the order of the group. But the order of the group in this example is $\infty$. I think we don't have a proposition which says that if $A^6 = 1$ then the order of $A$ divides $6$. –  meinzlein Feb 4 '12 at 11:52
@meinzlein: Suppose $x^k = 1$. Then $x$ has finite order, say $n$. By the division algorithm, $k = qn + r$, where $0 \leq r < n$. Thus $x^{qn + r} = 1$, and $x^r = 1$. Now $r$ cannot be positive since $r < n$. Thus $r = 0$ and $n$ divides $k$. No need for Lagrange here. In our case $x^6 = 1$ so the order of $x$ divides $6$. –  Mikko Korhonen Feb 4 '12 at 11:58

We can use the following relation for $2\times 2$ matrices: $$A^2-\operatorname{tr}(A)A+(\det A)I_2=0.$$ Here, it gives $A^2=A-I$, so $A^3=A^2-A=A-I_2-A=-I_2$ and $A^6=I_2$. So the subgroup generated by $\{A^k,k\in\mathbb N\}$ has at most $6$ elements. The order of $A$ divides $6$, $A^2\neq I_2$, $A^3\neq I_2$ so this order cannot be $2$ or $3$: it's necessarily $6$.