Vtyjkyuk

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 $mathbbR^3$ to $mathbbR^3$ with determinant $1$. Let $vinmathbbR^3$, $|v|=1$, and let $H_v=Tin Gmid Tv=v$.

(a) Show that $H_v$ is a subgroup of $G$.

(b) Let $S_v=Tin Gmid Ttext 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 $Ain G$ 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:

1. Take $Tin S_v$, 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 $v$.


2. In the post Prove that the set $fin mathfrak G$(where $w$ is a $r$th root of unity) is a coset of the subgroup $f(p)=1$., I got the statement that if $f$ is a group homomorphism and $bin operatornameImf$, 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 $AB in S_v$.