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Let $K'$ and $K''$ be two systems of coordinate axes obtained by application of a rotation $(n_1,\theta_1)$ followed by $(n_2,\theta_2)$
\begin{equation}
K\stackrel{(n_1,\theta_1)}{\longrightarrow} K'
\stackrel{(n_2,\theta_2)}{\longrightarrow} K''
\end{equation}
Let $x,x',x''$ denote components of position vector of a point x. with respect to the three sets of coordinates. Thus
\begin{equation}
x'=R_{\hat{n}_1}(\theta_1) x
\end{equation}
and
\begin{equation}
x''=R_{n_2}(\theta_2)x'
=R_{n_2}(\theta_2)R_{n_1}(\theta_1) x
\end{equation}
Thus
\begin{equation}
x''=R_{\hat{n_3}}(\theta_3) x
\end{equation}
where
\begin{equation}
R_{\hat{n_3}}=R_{n_2}R_{n_1}(\theta_1)
\end{equation}
is an orthogonal matrix with unit determinant and hence it corresponds to a rotation about an axis $\hat{n_3}$ by some angle $\theta$
\begin{equation}
K\longrightarrow K' \longrightarrow K''
\end{equation}
It is now straight forward to check that the set of all rotations form a group. This is most easily checked by noting that the set of all \(3\times 3\) matrices,
\begin{equation}
SO_{3}=\{R|R^T R=I, \text{det} R=+1\},
\end{equation}
is a group under matrix multiplication. This is the group \(3\times3\) special orthogonal matrices
In $SO_{3}$ is called special orthogonal group and $S$ refers to the special property of determinant begin unity.
The set of all orthogonal $N \times N$ matrices(with determinant +1 or -1) is called orthogonal group $0(N)$. The orthogonal group may also be defined as the set of all transformations $R$
\begin{equation}
x\longrightarrow x'=R x
\end{equation}
such that
\begin{equation}
x^{T} y=x'^T y'
\end{equation}
This set includes all proper $(\det R=+1)$ as well as improper rotations.$(\det R=-1)$
The group $S0(3)$ is a non Abelian group since the order of r combining two rotations is important when the axes are different
\begin{equation}
R_{\hat{n}_1}(\theta_1)R_{\hat{n}_2}(\theta_2)\ne
R_{\hat{n}_2}(\theta_2)R_{\hat{n}_3}(\theta_3)
\end{equation}
