# Showing that this set of functions is a group.

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?

• In addition to the excellent answers below, I will add that this group is isomorphic to $S_3$.
– user10575
May 4, 2013 at 11:01
• @Shalab Indeed, and this can be seen by how the functions act as permutations on the three-element set $\{0,1,\infty\}$ if we extend their definitions in the obvious manner. May 4, 2013 at 11:06
• Some post discussing that this group is isomorphic to $S_3$ can be found here and here. Oct 9, 2015 at 15:00

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 $g\circ f=f\circ g=f_1$

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 $X$.

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?

• This, together with the other answers, should be enough for me to try it again, as I think I might do better now, danke ;). May 4, 2013 at 11:16

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?

• Did you mean "only 6 functions... come on"? May 4, 2013 at 10:50
• After looking up "non-abelian group" I can confirm that this was a good hint! May 4, 2013 at 11:01
• No, it was "common", with a texan accent which I usually don't like but in this case fits precisely...:) May 4, 2013 at 11:02
• Oh, dear! You didn't know what an abelian-nonabelian group is yet you were given this exercise, @nullmoon ? That's odd...what course are you taking? May 4, 2013 at 11:03
• Well, I'm learnig a lot more here on Stackexchange then I do in lectures. Those exercises are made for groups, but I was left alone this week and now I have to deal with it. Course is called fundamentals of mathematics (if I translated correctly). May 4, 2013 at 11:11

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..

• "If f is in X, sending x to a, then the function g sending x to $\frac{1}{a}$ is the inverse of f" - this is not how inverse functions work; you're thinking about reciprocals (inverses under multiplication, not composition). "Actually f2 is the inverse of f1, f3 the inverse of $f_2$, $f_4$ of $f_3$ and so on..." -- this is also not true: in a group, the inverse of the identity is always the identity itself, so $f_2$ can't be the inverse of $f_1$. $f_2$ and $f_3$ are also their own inverses. The functions $f_1$, $f_2$, and $f_3$ correspond to the transpositions in the symmetric group $S_3$. May 4, 2013 at 11:16
• Not sure why I can't edit my comment, but I didn't mean to say $f_1$ is a transposition; it acts as the identity, not a transposition. May 4, 2013 at 11:24
• @symplectomorphic i was not speaking in general, i was referring to the actual case, where reciprocal and inverse coincide. your second remark is surely correct, i've edited now May 4, 2013 at 11:24
• but you're still wrong! reciprocation and inverses don't coincide. the inverse of $f_3$ is NOT $f_2$. these functions are called involutions -- they are obviously their own inverses. May 4, 2013 at 11:27
• @symplectomorphic no no, sorry, i was confused, you are absolutely right!! May 4, 2013 at 11:28

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.