Let $G=O(n)\times V^n$ and define a product in $G$ by $(A,v)(B,w)=(AB,Bv+w)$. Show that $\{I\}\times V^n$, $I$ being the identity matrix, is a closed submanifold and normal subgroup and that $O(n)\times \{0\}$ is a closed submanifold and subgroup $G$.
$O(n)$ stands for the orthogonal group.
I am seeing how to prove this because I cannot develop a map that would be a translation.
If $$F:G\to {I}\times V^n$$ such that $$F(g)= A^{-1}A +v=I+v$$.
$DF$ has full rank on every point, but I cannot find a regular point, that would give me the subgroup $H$.
Question: How should I prove the assertion?
Thanks in advance.
Your group $G=O(n)\times V^n$ is a trivial vector bundle over $O(n)$ with model fibre $V^n$. In particular, the projection $\pi_{O(n)}:O(n)\times V^n\rightarrow O(n)$ is a smooth submersion (this is true of any smooth fibre bundle, but in the case of trivial ones it's particularly easy to show this), so the inverse image, $\pi_{O(n)}^{-1}(g)$ of any $g\in O(n)$ is an embedded sub manifold. Furthermore since the singleton set $\{g\}$ is closed in $O(n)$, we have that $\pi_{O(n)}^{-1}(g)$ is closed in $O(n)\times V^n$. It is easy to see that $\pi_{O(n)}^{-1}(g)=\{g\}\times V^n$, so $\{g\}\times V^n$ is a closed embedded submanifold of $G$ for all $g\in O(n)$.
Showing that $\{I\}\times V^n$ is a subgroup of $G$ is easy, since $V^n$ is an abelian group by forgetting the vector space structure, and your group structure on the second coordinate is just vector addition. It follows by Cartans subgroup theorem that $\{I\}\times V^n$ is a closed Lie subgroup of $G$.
To show that $\{I\}$ is normal, let $h=(A,v)$, then $h^{-1}=(A^{-1}, A^{-1}(-v)$ as:
$$h\cdot h^{-1}=(A,v)\cdot (A^{-1},A^{-1}(-v)=(I,A^{-1}v-A^{-1}v)=(I,0) $$
Let $n=(I,w)$, then:
$$gng^{-1}=(A,v+w)(A^{-1},A^{-1}(-v)=(I,A^{-1}(v+w)-A^{-1}v)=(I,A^{-1}w)\in \{I\}\times V^n$$
so $G$ is a normal closed Lie subgroup of $G$.
You can do the same thing for your other case. In particular, $O(n)\times V^n$ is a trivial $O(n)$ bundle over $V^n$, and there is a smooth right action of $O(n)$ on $G$ given by $(A,v)\cdot B=(AB,v)$, which is free and transitive on $O(n)$, and preserves the fibers $\pi_{V^n}$. It follows that $O(n)\times V^n$ is a trivial principal $O(n)$ bundle over $V^n$, and everything we've done here works out again, in a probably even simpler way, except the subgroup isn't normal I don't believe.