I'm studying discrete math and in my book the definition of permutation is as follows :
Let $X$ be a non-empty set, it is called permutation of $X$ a bijective function $f: X \to X$.
The set of all permutations is denoted by $S_X$.
which I get , then it is written
To study $S_X$ where $|X| = n$ , it's enough to study $S_n$ (where $X = I_n = \{1,2,\dots,n\}$)
This is good to me because they have the same number of elements so it is intuitively true, but next there is the following proposition which should clarify the relation between $S_X$ and $S_n$ :
Let $X$ be a finite set with $|X| = n$. Then there's a biejection $f : S_X \to S_n$ so that for all $\sigma , \pi \in S_X$ it is true that $f(\sigma \circ \pi ) = f(\sigma) \circ f(\pi) $. In particular $|S_X |= n!$.
However I don't understand what does the part 'so that for all…$f(\sigma \circ \pi ) = f(\sigma) \circ f(\pi)$'. What's the intuitive and conceptual meaning for this formula? can you help me to understand this?
Thanks in advance
Best Answer
Here, $\circ$ denotes the composition of permutations, i.e. $\sigma \circ \pi$ is the permutation obtained by applying $\pi$, then applying $\sigma$. The statement says that there exists a "natural" bijection between $S_X$ and $S_n$, in the sense that it preserves the law of composition. If you have already heard about group morphisms, this just says that there exists a group isomorphism between $(S_X, \circ)$ and $(S_n, \circ)$.