Isometry can be written as a composition I need help proving:

Every isometry of $\mathbb{R}^n$ can be uniquely written as the
  composition $t\circ k$ where $t$ is a translation and $k$ is an
  isometry fixing the origin.

What do they exactly mean by "$k$ is an isometry fixing the origin" (I would like to know conceptually what they meant by that)? Thanks in advance! 
 A: We give a very formal proof of the fact that there is such a translation $t$ and isometry $A$. The argument is basically group-theoretic. 
Let $\phi$ be an isometry, and let $O$ be the origin. Suppose that $\phi(O)=A$. Let $t$ be the translation that takes the origin to $A$, and let $t^{-1}$ be the inverse of $t$. So $t^{-1}$ is the translation that takes $A$ to $O$.  Define $k$ by $k=t^{-1}\circ \phi$. 
It probably clear that $\phi=t\circ k$. For $t\circ k=t\circ(t^{-1}\circ \phi)=(t\circ t^{-1})\circ\phi=\phi$. 
Note that $k$ is an isometry, because it is a composition of two isometries. We check that if fixes the origin. To see this, apply $t^{-1}\circ\phi$ to $O$. We get $t^{-1}(\phi(O)$. But $\phi(O)=A$, and $t^{-1}(A)=O$. 
The problem also asked us to show that $t$ and $k$ are unique. So suppose that $\phi=t\circ k=t'\circ k'$, where $t$ and $t'$ are translations and $k$ and $k'$ fix the origin. We need to show that $t=t'$ and $k=k'$. 
Apply the transformation $t\circ k$ to the origin. We get $(t\circ k)(O)=t(k(O))=t(O)$. Similarly, $(t'\circ k')(O)=t'(O)$.
Thus $t(O)=t'(O)$. So the two translations $t$ and $t'$ do the same thing to a certain point $O$, and therefore they are the same.  Then from $t\circ k=t'\circ k'$ we conclude that $k=k'$.
