I solved the following exercise. Could someone please check my work?
Prove that $\operatorname{Isom_n}{\mathbb R^n}$ is a matrix group. Is it compact?
My work:
We recall that
$$\operatorname{Isom_n}{\mathbb R^n} = \left \{ \left ( \begin{array}{cc} A & 0 \\ V & 1 \end{array} \right ) \mid A \in O(n) (\mathbb K), V \in \mathbb R^n \right \} $$
By considering blockwise multiplication, blockwise inversion and the multiplicativity of the determinant it is clear that $\operatorname{Isom_n}{\mathbb R^n} $ is a subgroup of $GL_{n+1}$.
Since $|\det|$ is continuous and
$$ \det \left ( \begin{array}{cc} A & 0 \\ V & 1 \end{array} \right ) = \det A$$
it is clear that if $A_n \to A$ and $|\det A_n| = 1$ then also $|\det A| = 1$ hence $\operatorname{Isom_n}{\mathbb R^n} $ is closed in $GL_{n+1}$.
It is however not closed in $M_{n+1}$. To see this note that
$$ X_n = \left ( \begin{array}{cc} D_n & 0 \\ V & 1 \end{array} \right ) $$
where $D_n$ is the diagonal matrix with $d_1 = {1\over n}$, $d_2 = n$ and $d_k = 1$ otherwise is a sequece in $\operatorname{Isom_n}{\mathbb R^n}$ that converges to a matrix of zero determinant.
Similarly we see that $\operatorname{Isom_n}{\mathbb R^n}$ is not compact: the norm of $X_n$ is not bounded hence $\operatorname{Isom_n}{\mathbb R^n}$ is not bounded and therefore not compact.
The question doesn't reduce to just showing that $O(n)$ is closed in $GL_n(\mathbb{R})$. In the formula for $Isom_n(\mathbb{R}^n)$, you have not only an $A \in O(n)$, but also $V \in \mathbb{R}^n$, which you never considered.