# Prove that, for any set $X$, $\mathrm{S}X$ is a group under function composition

• Given a set $$X$$, a permutation of $$X$$ is a bijection $$\sigma : X \longrightarrow X$$. That is, it is a function with domain and codomain $$X$$ which is one to one and onto.

• Given a set $$X$$, the set $$\mathrm{S}X$$ is the set of all permutations of $$X$$.

Prove that, for any set $$X$$, $$\mathrm{S}X$$ is a group under function composition.

I understand the four properties it needs to prove that it's a group but I don't know how to apply it to this question.

• Can you so say the identity property? Sep 2, 2019 at 18:23
• Try to prove them one by one? Do you have any guess for the permutation playing the role of neutral element, for example? Sep 2, 2019 at 18:23
• If $\sigma$ and $\tau$ are bijections from $X$ to $X$, is $\tau\circ \sigma$ a bijection from $X$ to $X$ [closure]? If $\sigma,\tau,\rho$ are bijections, does $(\rho\circ\tau)\circ \sigma$ equal $\rho\circ(\tau\circ\sigma)$ [associativity]? Is there a bijection $\iota\colon X\to X$ such that for all bijections $\sigma$, $\iota\circ\sigma=\sigma\circ\iota=\sigma$ [identity]? Who? Given $\sigma$, does there exist a bijection $\tau$ such that $\sigma\circ \tau = \tau\circ\sigma=\iota$ (from previous property) [inverses]? You must answer each affirmatively and prove this is the case. Sep 2, 2019 at 19:19
• Incidentally, I think historically things went just the other way around: the "four properties" of abstract groups were patterned just upon the four properties holding for the bijections recalled in the previous comment.
– user615081
Sep 3, 2019 at 15:21