1. ## Group theory

Having some trouble figuring this out - any help at all would be greatly appreciated! Thanks

S the group of bijective applications R - > R. x, y elements of S. For every real number t, x(t) = t +1, y(t) = 2t.
G is a sub-group of S generated by {x,y}.

1. For every integer n>=1, x
n = y^(-n)xy^(n).
Calculate x
n(t) and
show that (x
n)^2 = xn-1

2. Let Hn be a subgroup of G generated by xn Show that Hn is a subgroup of Hn+1 and that Hn is not equal to Hn+1

2. Originally Posted by rmangan

S the group of bijective applications R - > R. x, y elements of S. For every real number t, x(t) = t +1, y(t) = 2t.
G is a sub-group of S generated by {x,y}.

1. For every integer n>=1, x
n = y^(-n)xy^(n).
Calculate x
n(t) and
show that (x
n)^2 = xn-1

2. Let Hn be a subgroup of G generated by xn Show that Hn is a subgroup of Hn+1 and that Hn is not equal to Hn+1
The elements $\displaystyle x,y$ of all bijections on $\displaystyle \mathbb{R}$ are functions $\displaystyle x,y:\mathbb{R}\to \mathbb{R}$ defined by $\displaystyle x(t) = t+1$ and $\displaystyle y(t) = 2t$. Now $\displaystyle x^n(t) = x\circ x \circ ... \circ x (t) = (((t+1)+1)+1) + ... + 1 = t+n$. Also, $\displaystyle y^n (t) = y\circ ... \circ y(t) = 2(...(2(2t))) = 2^n t$.
Thus, $\displaystyle x_n(t) = y^{-n} x y^{n} (t) = 2^{-n}(2^{n} t + 1) = t + 2^{-n}$.
We see that $\displaystyle x_{n}^2 (t) = (t+2^{-n})+2^{-n} = t + 2\cdot 2^{-n} = t + 2^{1-n}$ and so $\displaystyle x_n = x_{n-1}^2$.

Wii see that $\displaystyle \left< x_n \right> \subseteq \left< x_{n+1} \right>$ because $\displaystyle x_n = x^2_{n+1}$.
Thus, $\displaystyle H_n \subseteq H_{n+1}$ and can you show now $\displaystyle H_n\not = H_{n+1}$?