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Orhogonal group O is a group. How is $O(3)/O(2)$ partitioned???

$O(n)=\{ A \in Gl(n):A^t=A^{-1}\}$

$O(3)/O(2)$ is supposed to be a quotient group where $O(2)$ is normal.

so $O(3)/O(2)$ is a way to partitioned $O(3)$?.I trying to use cosets

$g\in O(3)$ $g*O(2)$ is supposed to be a partition but Cant multiplied them. Error Dimensions don't agree. Guessing O(2) is within a 3x3 matrix? can it be broken down??

Tiger Blood
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  • I don't agree with the proposal to close your question but, for next time, pay attention to the way you write, for example your title is almost meaningless, "I trying to use" $\to$ "I have been trying to use", etc. – Jean Marie Sep 20 '17 at 18:39

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There isn't a unique copy of $O(2)$ inside of $O(3)$. You're free to pick the copy of $O(2)$ to be able to compute the cosets.

You should probably just pick it to be the group generated by

$\begin{bmatrix}\cos(\theta)&-\sin(\theta)&0\\\sin(\theta)&\cos(\theta)&0\\0&0&1\end{bmatrix}$ and

$\begin{bmatrix}\sin(\theta)&\cos(\theta)&0\\\cos(\theta)&-\sin(\theta)&0\\0&0&1\end{bmatrix}$

Then you'll be able to compute the set of cosets. However, $O(2)$ isn't a normal subgroup of $O(3)$, so you cannot hope for a quotient group.

rschwieb
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  • @Ugo there is a "vision" of $O(3)/O(2)$ as a particular Stiefel manifold (https://math.stackexchange.com/q/75024). – Jean Marie Sep 20 '17 at 18:33