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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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    @meinzlein: $1 = x^{qn+r} = x^{qn}x^r = (x^n)^qx^r = 1^qx^r = x^r$ because $n$ is the order of $x$.2012-02-04

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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 $6$.