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$S_n$ is the set of all permutations.

I'm just starting on this material, so I'm confused on how to read this problem. Does the function consist of multiple permutations (i.e. the permutation of a permutation)?

A property of a permutation of $\{1, ..., n\}$ is that it is a bijection to itself. So does this property automatically make $C_π$ a bijection?

Chris
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2 Answers2

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The easiest way to show that $C_{\pi}$ is a bijection is to exhibit an inverse. And if $\pi^{-1}$ is the inverse of $\pi$, then $$ C_{\pi^{-1}}(C_{\pi}(\sigma))=\pi^{-1}\pi\sigma=\sigma $$ and similarly $C_{\pi}(C_{\pi^{-1}}(\sigma))=\sigma$.

Therefore $C_{\pi}$ has inverse $C_{\pi^{-1}}$.

This is a special case of a general property: when a group $G$ acts on a set $X$, the function $f_g:X\to X$ induced by the action of $g\in G$ is always a bijection.

carmichael561
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If you can assume that

  • $\forall g \in S_n, g$ has an inverse,

  • $S_n$ is closed under composition, and

  • composition of permutations are associative,

Suppose $C_\pi(a) = C_\pi(b)$, then

$\pi a = \pi b \implies a = b \quad$ (since $\pi^{-1}$ exists and the operation is associative)

So $C_\pi$ is injective.

Now take an arbitrary $c \in S_n$, is there an $x \in S_n$ such that $C_\pi(x) = c$ ?

Yes, that element is $\pi^{-1}c,$ (by assumption $\pi^{-1}c \in S_n$ by closure) since

$C_\pi(\pi^{-1}c) = \pi\pi^{-1}c = c \quad$ (since $\pi^{-1}$ exists and composition is associative)

So $C_\pi$ is surjective.

$\therefore C_\pi$ is bijective

crackpotHouseplant
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