Question: Let $S$ be a nonempty set and $A(S)$ be the set of all one-to-one mappings of $S$ onto itself. I.N. Herstein in *Topics in Algebra* says (in page 28) that whenever $S$ has three or more elements, we can find two elements $\alpha,\beta\in A(S)$ such that $\alpha \circ \beta \neq \beta \circ \alpha$.

Why is this true? Suppose we assume on the contrary that for every $\alpha,\beta\in A(S)$, we have $\alpha \circ \beta = \beta \circ \alpha$, how do we obtain a contradiction to this assumption?

Although <https://math.stackexchange.com/a/1058719/114758> gives an example (and hence proof), is there a way to contradict my assumption by not taking an example?

Answer: 
There is a way to do what you are saying, but based on your demand, the proof is slightly more lengthy, and involves more terminology than you would like.

So we know that the set of one-to-one mappings from $S$ to itself, form a group under composition. Note that this group is abelian, because of our assumption. Further, the order of this group is $3!=6$.

So it is known that our group is abelian. Consider the conjugacy classes of the elements of $S\_3$. By considering the cycle type, you see that these are in bijection with the partitions of $n$. It is known that an element is in the center if and only if it's conjugacy class has only one element. That way we would be expecting $6$ conjugacy classes if the group were abelian, because then the center would be the whole group, however the number of partitions of $3$ is $4$, because $3=3=2+1=1+2=1+1+1$, so there are four representations. This is a proof, without exhibition of an example, that $S\_3$ is non-abelian. For $S\_n$, find yourself a way of proving that $p(n)<n!$ (it's easy actually) so that $S\_n$ will also be non-abelian.

An even stronger theorem is that the center of $S\_n$ is trivial, that is to say, other than the identity, **no other bijection commutes with all other bijections**.

So this is a proof the way you like it, albeit technical. Please ask if you have further doubts or would like me to explain the cyclic decomposition.
Take $\sigma\_1 = (1 \ 2 \ 3), \sigma\_2 = (1\ 2)$ then
$\sigma\_1 \circ \sigma\_2 = (1 \ 3)$ and $\sigma\_2 \circ \sigma\_1 = (2 \ 3)$.

Now use this to generate a contradiction.