1. ## Group theory

Hey!

For any subset ¬ of the affine plane R^2, let G(¬) denote the group of all affine transformations f of R2 such that f(¬) = ¬.

The elements of the group G = G(¬), where ¬ is the hyperbola xy = 1
in R^2; i think are: (ax|dy) & (by|cx) where ad=bc=1

Show that the elements of G preserving each of the two asymptotes of ¬ form a normal subgroup N of index 2 in G, isomorphic to the multiplicative group R* of non-zero real numbers, which acts regularly on ¬.

Do all the elements of G then preserve the two asymptotes? Any hints how to show the rest please?!

No idea what i'm doin really i'm afraid, but here goes nothing:

The index of a subgroup N of G is the number of distinct left
(or right) cosets of N in G. Even if N is not a normal subgroup, the corre-
spondence xN -> Nx is a bijection between the set of left cosets and the
set of right cosets. If the index of N in G is equal to 2, then N is a normal subgroup of G:

Showing that Nx = xN for each x in G. If x in N there is nothing
to prove. Hence, let x is not in N. Since the index is two, the disjoint union of N and xN make up all of G. Analogously the disjoint union of N & Nx = G.
Since N does not equal Nx as well as N does not equal xN, it follows that Nx = xN.
Multiplying this with x^-1 gives that N^x = N and hence N is a normal
subgroup.

???
Thanks, any help would be much appreciated. x

2. Originally Posted by dazed
Hey!

For any subset ¬ of the affine plane R^2, let G(¬) denote the group of all affine transformations f of R2 such that f(¬) = ¬.

The elements of the group G = G(¬), where ¬ is the hyperbola xy = 1
in R^2; i think are: (ax|dy) & (by|cx) where ad=bc=1

Show that the elements of G preserving each of the two asymptotes of ¬ form a normal subgroup N of index 2 in G, isomorphic to the multiplicative group R* of non-zero real numbers, which acts regularly on ¬.

Do all the elements of G then preserve the two asymptotes? Any hints how to show the rest please?!

No idea what i'm doin really i'm afraid, but here goes nothing:

The index of a subgroup N of G is the number of distinct left
(or right) cosets of N in G. Even if N is not a normal subgroup, the corre-
spondence xN -> Nx is a bijection between the set of left cosets and the
set of right cosets. If the index of N in G is equal to 2, then N is a normal subgroup of G:

Showing that Nx = xN for each x in G. If x in N there is nothing
to prove. Hence, let x is not in N. Since the index is two, the disjoint union of N and xN make up all of G. Analogously the disjoint union of N & Nx = G.
Since N does not equal Nx as well as N does not equal xN, it follows that Nx = xN.
Multiplying this with x^-1 gives that N^x = N and hence N is a normal
subgroup.

???
Thanks, any help would be much appreciated. x
identify $\displaystyle G$ with $\displaystyle \left \{\begin{pmatrix}a & 0 \\ 0 & \frac{1}{a} \end{pmatrix}: \ a \in \mathbb{R}^{\times} \right \} \bigcup \left \{\begin{pmatrix}0 & a \\ \frac{1}{a} & \ 0 \end{pmatrix}: \ a \in \mathbb{R}^{\times} \right \}$ and $\displaystyle N$ by $\displaystyle \left \{\begin{pmatrix}a & 0 \\ 0 & \frac{1}{a} \end{pmatrix}: \ a \in \mathbb{R}^{\times} \right \}.$ looking at $\displaystyle H=\{1,-1 \}$ as a multiplicative subgroup of $\displaystyle \mathbb{R},$ we define $\displaystyle f: G \longrightarrow H$

by $\displaystyle f(A)= \det A.$ see that $\displaystyle f$ is a surjective group homomorphism and $\displaystyle \ker f = N.$ thus $\displaystyle N$ is a normal subgroup of $\displaystyle G$ and since $\displaystyle G/N \cong H$ and $\displaystyle |H|=2,$ the index of $\displaystyle N$ in $\displaystyle G$ is $\displaystyle 2$.

the isomorphism between $\displaystyle N$ and $\displaystyle \mathbb{R}^{\times}$ is just the obvious one: $\displaystyle \begin{pmatrix}a & 0 \\ 0 & \frac{1}{a} \end{pmatrix} \mapsto a.$