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 $id\circ\alpha=\alpha\circ id=\alpha$


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 $\sigma\gamma=\gamma\sigma$?

  • $\begingroup$ Possible duplicate of Prove that the symmetric group $S_n$, $n \geq 3$, has trivial center. $\endgroup$ – Dietrich Burde Oct 15 '17 at 13:48
  • $\begingroup$ @DietrichBurde It seems to me that the question here is different. The OP seems interested not in the answer to the problem (and in fact they know the answer) by where their thinking about the problem went wrong. $\endgroup$ – Stella Biderman Oct 15 '17 at 14:50
  • $\begingroup$ The thinking here "which went wrong" is simply that it is not about commuting $\alpha\circ \beta=\beta\circ \alpha$ of single elements, but $\sigma\circ \gamma=\gamma\circ \sigma$ for all $\gamma\in S_n$. This is explained in the the duplicate, too. $\endgroup$ – Dietrich Burde Oct 15 '17 at 14:54

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.


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