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..

Answer 4

Since $f_1$ is the identity function, you immediately see that $f_1\circ f_i=f_1\circ f_1=f_i$.

You can see that $f_2\circ f_2=id=f_1$ and $f_3\circ f_3=id=f_1$. It is also easy to see that $f_3\circ f_2=f_4$ and $f_2\circ f_3=f_5$. (Simply calculate those compositions.)

So far you have $$ \begin{array}{c|cccccc} \circ & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\\hline f_1 & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\ f_2 & f_2 & f_1 & f_5 & & & \\ f_3 & f_3 & f_4 & f_1 & & & \\ f_4 & f_4 & & & & & \\ f_5 & f_5 & & & & & \\ f_6 & f_6 & & & & & \end{array} $$ Using the equations you already have, you should be able to find, $f_4\circ f_5$ and $f_5\circ f_4$. (Just try $f_4\circ f_5=f_3\circ f_2\circ f_2\circ f_3=\dots$ Notice that you are using associativity of the composition here.)

Are there some more spots which you can fill in using the equations you already know? If not take some other two function, calculate the composition and try to continue in this way.