I am stuck while solving the following problem:
Problem (Source: University of California, Berkeley - Prelim exams)
Let $G$ be the group of orthogonal transformations of $\mathbb{R}^3$ to $\mathbb{R}^3$ with determinant $1$. Let $v\in\mathbb{R}^3$, $|v|=1$, and let $H_v=\{T\in G\mid Tv=v\}$.
(a) Show that $H_v$ is a subgroup of $G$.
(b) Let $S_v=\{T\in G\mid T\text{ is a rotation through $180^\circ$ about a line orthogonal to $v$}\}$. Show that $S_v$ is a coset of $H_v$ in $G$.
I easily solved step (a), so I know that $H_v$ is a subgroup of $G$. If I find a matrix $A\in G$ such that $S_v=AH_v$, then I succeed to show that $S_v$ is a coset of $H_v$ in $G$. However, I don't know how to find $A$.
My notes:
- Take $T\in S_v$, then $T^2$ is a rotation through $360^\circ$ about a line, so $T^2$ is the identity. Especially, $Tv=-v$. Thus $T$ have an eigenvalue $-1$ associated with $v$, while elements in $H_v$ have an eigenvalue $1$ associated with $v$.
- In the post Prove that the set $\{f\in \mathfrak G|f(p)=w\}$(where $w$ is a $r$th root of unity) is a coset of the subgroup $\{f\in \mathfrak G|f(p)=1\}$., I got the statement that if $f$ is a group homomorphism and $b\in \operatorname{Im}f$, then $f^{-1}(b)$ is a coset of $\ker f$.
Hint, since you say you don't know how to find $A$: $H_v$ is a subgroup, so $I \in H_v$ (where $I$ is the identity matrix). You're trying to find $A$ such that $S_v = AH_v$; since $I \in H_v$, this implies $A \in S_v$. (And in fact, you can take any such $A$). Therefore, what you want to show is that given any $A \in S_v$, $B \in H_v$, then $AB \in S_v$.