My problem seems somewhat basic yet I can't seem to solve it directly. The problem is this:
Suppose $\mathbf{x}\in \mathbb{R}^n$ is orthogonally invariant, i.e., $$ \mathbf{x} = \mathbf{H}\mathbf{x},\ \forall\, \mathbf{H}\in \mathbf{O}_n $$ where $\mathbf{O}_n = \{\mathbf{K}\in \mathbb{R}^{n\times n}: \mathbf{K'K}= \mathbf{KK'} = \mathbf{I}_n\}$. If $\mathbf{x} = (\mathbf{x}_1', \mathbf{x}_2')'$ is some ''partition'' of $\mathbf{x}$, show that $\mathbf{x}_1$ is also orthogonally invariant.
I attempted partitioning $\mathbf{H}$ in my direct proof but it utterly failed. I was aiming to arrive with something like $\mathbf{x}_1 = \mathbf{H}_1\mathbf{x}_1,\ \forall\, \mathbf{H}_1\in \mathbf{O}_m,\ m<n$. Any help? Preferably, I am looking for a non-abstract algebra solution.
Suppose that the partition includes the first $m$ coordinates, i.e. using the splitting $\mathbb R^n=\mathbb R^m\times \mathbb R^{n-m}$. Then every $O_m$ can be embedded into an $O_n$ by forming the block diagonal matrix $$ \begin{pmatrix} O_m&0\\0&I_{n-m} \end{pmatrix} $$ which is easily seen to be orthogonal. Since this matrix acts like $O_m$ on the first $m$ coordinates and it is the identity on the last $n-m$ coordinates, it follows that $x_1$ is orthogonally invariant.