The transformation group of all bijections has a subset V of all strictly increasing bijections, show its a subgroup. Question: 
Consider the transformation group $S_{\rm I\!R}$ of all bijection's from $\rm I\!R$ $\longrightarrow$ $\rm I\!R$. Show that V$\subset$ $S_{\rm I\!R}$ consisting of all strictly increasing bijections forms a subgroup of $S_{\rm I\!R}$. Is it a Normal subgroup?
So I know the basics of group theory, further I know the criteria for being a subgroup are that V should be stable (closed under composition) and that the inverse of each element should also be in V. 
So I should show that firstly, V is a subgroup. I have no idea how though. I assume V is closed under composition and associative since bijections are one to one correspondence, also therefore invertible, so the inverse exists. Identity I'm not sure about but if the inverse exists by definition, composition of an element with its inverse implies the neutral element exists also. Right? I'm just thrown by the idea of a strictly increasing bijection, I don't see how this affects things.
 A: Actually, since $S_{\mathbb R}$ is a group, then, in order to prove that $V$ is a subgroup of $S_{\mathbb R}$ you only have to check two things:


*

*$f,g\in V\implies g\circ f\in V$;

*$f\in V\implies f^{-1}\in V$.


Both of them are clear: if you compose two strictly increasing functions, then you get again a strictly increasing function and the inverse of a strictly increasing function is again a strictly increasing function.
However, it is not a normal subgroup. Take, for instance,$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}-1&\text{ if }x=1\\1&\text{ if }x=-1\\x&\text{ otherwise}\end{cases}\end{array}$$and $g(x)=x+2$. Then $g\in V$. But $f^{-1}\circ g\circ f\notin V$, since $(f^{-1}\circ g\circ f)(1)=-1$ and $(f^{-1}\circ g\circ f)(-1)=3$. So, $(f^{-1}\circ g\circ f)(-1)>(f^{-1}\circ g\circ f)(1)$, in spite of the fact that $-1<1$
A: Well, V is the subset of bijections $f$ of the real line for which $x > y$ implies $f(x) > f(y)$. For being a subgroup you have to check that 
1) It's closed under composition: Say $f,g \in V$, $x > y$, then $g(x) > g(y)$, and consequently $(f \circ g)(x) = f(g(x)) > f(g(y)) = (f \circ g)(y)$.
2) It contains Inverses: But that's also obvious, because $x >y \iff f(x) > f(y)$.
To check that it is a normal subgroup, you'll need to conjugate $f \in V$ by ANY $g \in S_\mathbb{R}$ (i.e. $g \circ f \circ g^{-1}$), and see if that is still increasing. You can convince yourself that it isn't by just some concrete example: Say $f(x) = x+1$, and g is such that $g^{-1}(1) = 0, g(1) = 1000, g^{-1}(2)=3, g(4)= -1000$ (remember that a bijection can be very far from being continuous). Now what's $(g \circ f \circ g^{-1})(x)$ for $x=1,2$? 
