Krdytkyu

I found the following result:

If $V$ is a euclidean vector space of finite dimension $n$ and $B$ is an orthonormal basis of $V$, then an endomorphism $f:Vto V$ is an isometry iff its matrix respect to $B$ is orthogonal.

The proof is not hard. If $f$ is an isometry and $A$ is its matrix respect to $B$, then
begin{equation*}
X^top mathbb{I} Y =langle x, y rangle =langle f(x),f(y)rangle =(AX)^top mathbb{I} (AY)=X^top A^top A Y,
end{equation*}
where $X$ and $Y$ are the coordinates of the vectors $x$ and $y$ with respect to $B$, respectively. As the vectors were arbitrarily chosen, it follows that $A^top A=mathbb{I}$, so $A$ is orthonormal. The reciprocal statement is obvious from this.

Then, I thought that the orthogonal group can be equal to the isometry group, but I read that this is not true: the isometry group has order $n(n+1)/2$ (because include the traslations) and the orthogonal group has order $n(n-1)/2$. However, I cannot see why is this way.

Translations aren't linear (they don't take $0$ to $0$); but are isometries. There are $n$ dimensions of translations in dimension $n$, accounting for the difference.