How associativity holds on set of all bijective functions on a finite set?

Question

I came across following statement:

The set of all bijective functions on a finite set forms a group under function composition.

The proof was given as follows:

We can have identity function as $\{(1,1),(2,2),(3,3),(4,4)\}$.
Since function is bijective and mapping to same set, we can have an inverse for any function by inverting the relation (changing the mapping $a\rightarrow b$ to $b\rightarrow a$).
Since the function maps to the same set, it must be closed and associative also.

I was thinking is associativity really holds in above set. For example, consider the finite set: $\{1,2,3,4\}$ and two bijective functions $f_1$ and $f_2$ as follows:

Now, $f_1(f_2(1))=f_1(3)=3$ and $f_2(f_1(1))=f_2(2)=1$. So, I feel its not bijective. Am I incorrectly evaluating associativity?

Answer 1

Your example shows that compostion is not commutative. It has nothing to do with being associative.

Answer 2

If we assume $f_1(a) = b$ and $f_2(b) = c$ and $f_3(c)= d$ then

$f_2\circ f_1(a) = f_2(f_1(a)) =f_2(b) =c$.

And $f_3\circ f_2(b) = f_3(f_2(b)) =f_3(c)=d$.

And so

$f_3\circ (f_2\circ f_1)(a) = f_3(f_2\circ f_1(a))=f_3(c) =d$

and $(f_3\circ f_2)\circ f_1(a) = f_3\circ f_2(f_1(a)) =f_3\circ f_2(b) = d =f_3\circ (f_2\circ f_1)(a)$

So associativity holds.