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With curiousity, I'm trying to prove whether multiplication of all rotation matrixes in $\mathrm{SO}(3)$ is identity irrelevant of multiplication order. As each rotation matrix in $\mathrm{SO}(3)$ space has an inverse rotation matrix in that space, if I multiply them ordered as $$R_1*R_1'*R_2*R_2'*....R_n*R_n' = I $$ would result in identity. However, if the multiplication order is random, is it possible that the result will be identity?

I tried to prove this in 2D rotation matrices. But because there is only one angle, it seems rotation operation has commutation property in this case. However in $\mathrm{SO}(3)$, rotation using three angles does not have commutation.

In $\mathrm{SO}(3)$, I tried taking logarithm of both sides and writing the multiplication as summation. Then I tried to convert it to triple integrals (for each infinitesimal angle) but I could not go on further from there.

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    The way you write your product (with the $n$ at the end), you seem to under the impression that the "product" of all the elements in $\mathrm{SO}(3)$ is finite. It's not. There a infinitely many elements in $\mathrm{SO}(3)$. So your first question should be what the product of all the rotations even *means*.2012-12-06
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    @user637387: the question you are asking, at least the way you phrased it, is similar to asking yourself what the product of all real numbers is.2012-12-06
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    @MartinArgerami Although the answer to your question is quite a bit easier--it's obviously zero :D2012-12-06
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    Yeah, I thought about that. But I forgot to add "nonzero".2012-12-06

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