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Let $n\geqslant 3$. Show that the unique element $\sigma$ of $S_n$ that satisfies $\sigma\gamma=\gamma\sigma$ for all of $\gamma\in S_n$ is the identity(id.)

Prepostion: If $\alpha,\beta$ are disjoint, they commute.

If we have $\alpha\in S_n$, then $id\circ\alpha=\alpha\circ id=\alpha$

Question:

I am not understanding what is asked. If the permutation are disjoint or if they are elevated to a certain power(ex:$\alpha^2=\alpha\circ\alpha)$, they commute. How can id(identity) be the only element that assures $\sigma\gamma=\gamma\sigma$?

Pedro Gomes
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1 Answers1

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What you're missing is that $\sigma$ needs to commute with every element of $S_n$. Yes, it is certainly the case that there exist pairs $\sigma_1,\sigma_2\in S_n$ such that $\sigma_1\sigma_2=\sigma_2\sigma_1$. However, given such pairs you can (as long as neither $\sigma_i$ is the identity) always find another permutation $\tau\in S_n$ such that $\sigma_1\tau\neq \tau\sigma_1$ (or $\sigma_2\tau\neq \tau\sigma_2$). The problem is asserting that the only permutation that commutes with every element of $S_n$ is the identity.

We can write this more formally as

$$(\forall\sigma\in S_n)((\forall\tau\in S_n,\sigma\tau=\tau\sigma)\Rightarrow\sigma=id_{S_n})$$

If you're interested in a proof of the theorem, you can find it here.