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Could someone explain the notation in this question to me so I can have a go at answering it.

Show that $SO_3(\mathbb{F_2}) = \{M \in SL_3(\mathbb{F_2})|M^{-1} = M^t\}$, where $M^t$ is the transpose of $M$.

Obviously $M$ is some matrix but I don't understand $SO_3(\mathbb{F_2})$, I think it has something to do with fields which I have next to zero experience with, although I presume very little experience with them is required for this question or we wouldn't have been given it.

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Google "special orthogonal group". –  Matt N. Nov 13 '12 at 20:23
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To me, that is the definition of $SO_3(\mathbb{F}_2)$. Your book or notes must include some alternative definition, since your job here is to show that the two definitions are equivalent. I would search your book and/or notes for the definition they want you to use for this problem. –  Brett Frankel Nov 13 '12 at 20:37

1 Answer 1

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In general, the special orthogonal group over a given field is formed by the orthogonal matrices with determinant $1$. Since in $\mathbb F_2$ the only possible determinant for an invertible matrix is $1$, the only condition for a matrix to be in the special orthogonal group is that the matrix be orthogonal.

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