• 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.

  • $\begingroup$ Can you so say the identity property? $\endgroup$ Sep 2, 2019 at 18:23
  • $\begingroup$ Try to prove them one by one? Do you have any guess for the permutation playing the role of neutral element, for example? $\endgroup$ Sep 2, 2019 at 18:23
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 2, 2019 at 19:19
  • $\begingroup$ 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. $\endgroup$
    – user615081
    Sep 3, 2019 at 15:21


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