Showing that this set of functions is a group.

Question

I have trouble understanding the following task:

Show that the set $X = \{f_1,\ldots ,f_6\}$ of functions $f_i : \Bbb Q\setminus \{0,1\} \to \Bbb Q\setminus \{0,1\}$ with
$x ↦ f_1(x) = x$,
$x ↦ f_2(x) = 1/x$,
$ x ↦ f_3(x) = 1-x,$
$x ↦ f_4(x) = 1/(1-x)$,
$ x ↦ f_5(x) = (x-1)/x$,
$x ↦ f_6(x) = x/(x-1)$
and the composition as the operation is a group. Is this group commutative?

I'm a little confused as I don't know what to do with all these functions. I can't apply the definitions of a group on this because of that. If the group is $(X,∘)$ how do I show that this is associative don't I need the information on how the single functions are connected or am I missing something?

Answer 1

First of all, composition of functions is always associative (in a way this is "the mother of all associativities"), because for all $x$ $$ ((f\circ g)\circ h)(x)=(f\circ g)(h(x))=f(g(h(x)))=f((g\circ h)(x))=(f\circ(g\circ h))(x).$$

Thus, what you need to show about your set $X$ of functions is

• if $f,g\in X$, then $f\circ g$ in $X$.
• there is a neutral element (which is $f_1$, obviously)
• for each $f\in X$ there is an inverse element, i..e some $g$ with $g\circ f=f\circ g=f_1$

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One can also try this the other way around: The set of bijections $\mathbb Q\setminus \{0,1\}\to\mathbb Q\setminus \{0,1\}$ is certainly a group $G$. If we start with $f_2,f_3\in G$, there is certainly a smallest subgroup $\langle f_2,f_3\rangle$ of $G$ that contains both $f_2$ and $f_3$. To find this group (noting that $f_2, f_3$ are their own inverses), we can simply keep playing with composing found functions with $f_2$ and $f_3$ until we no longer find new functions - a process that fortunately terminates and gives us exactly the set $X$.

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Another interesting point to note is that all elements of $X$ have the form $f(x)=\frac{ax+b}{cx+d}$ with $ad-bc\ne0$ (where we ignore the fact that $f$ is not defined at $x-\frac dc$, say we view these as functions on $\mathbb R\setminus \mathbb Q$). This reminds us of invertible matrices $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ with $\det A=ad-bc\ne0$. Do you see how the composition of two fucntions of this "matrix type" relates to a well-known operation among the corrsponding matrices?

Answer 2

You'll need to work a little hard, but not too much (only 6 functions...common). For example:

$$f_3\circ f_4(x):=f_3\left(\frac1{1-x}\right)=1-\frac1{1-x}=-\frac x{1-x}=\frac x{x-1}=f_6(x)$$

$$f_4\circ f_3(x)=f_4(1-x)=\frac1{1-(1-x)}=\frac1x=f_2(x)$$

The above, besides showing closedness in two particular cases, also shows that if this is a group then it is a non-abelian one of order $\,6\;$...this is a good hint, isn't it?

Answer 3

Composition of functions of any kind is always associative, so you have nothing to check. Closure under composition is immediate, take for example $$f_2\circ f_3:x\rightarrow 1-x\rightarrow \frac{1}{1-x}=f_4(x)$$

The identity function is in $X$, hence you have also a neutral element. As for the inverse, $f_1$ is its own inverse, being the identity, $f_2,f_3,f_6$ are again their own inverses, as you can check computing their squares, finally $f_4,f_5$ are one the inverse of the other..