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Compute the order of each element in the general linear group of dimension 2 over Z/(2)


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.4 Exercise 1.4.2

Write out all the elements in $GL_2(\mathbb{F}_2)$ and compute the order of each element.


Solution: We found the elements of $GL_2(\mathbb{F}_2)$ in a previous example.

$A_1 = \left[ {1 \atop 0}{0 \atop 1} \right]$ is the identity matrix, so $|A_1| = 1$.

$A_2 = \left[ {1 \atop 0}{1 \atop 1} \right]$: We have $A_2^2 = I$, so $|A_2| = 2$.

$A_3 = \left[ {1 \atop 1}{0 \atop 1} \right]$: We have $A_3^2 = I$, so $|A_3| = 2$.

$A_4 = \left[ {0 \atop 1}{1 \atop 0} \right]$: We have $A_4^2 = I$, so $|A_4| = 2$.

$A_5 = \left[ {1 \atop 1}{1 \atop 0} \right]$: We have $A_5^2 = \left[ {0 \atop 1}{1 \atop 0} \right]$, and $A_5^3 = I$, so $|A_5| = 3$.

$A_6 = \left[ {0 \atop 1}{1 \atop 1} \right]$: We have $A_6^2 = \left[ {1 \atop 1}{1 \atop 0} \right]$, and $A_6^3 = I$, so $|A_6| = 3$.


Linearity

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This Post Has 2 Comments

  1. Can you explain how you got $A_2^2 = A_3^2 = A_5^3 = A_6^3 = I$?

    For example, I get $A_2^2 = \begin{bmatrix} 1 & 2 \\ 0 &1 \end{bmatrix}$.

    Am I missing something obvious?

    1. Oh I understand, we're working with the field $\mathbb{F}_2$...

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