The Center of the Symmetric group is Trivial if $n>2$

Group Theory Problems and Solutions

Problem 31

Show that the center $Z(S_n)$ of the symmetric group with $n \geq 3$ is trivial.

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Steps/Hint

  1. Assume $Z(S_n)$ has a non-identity element $\sigma$.
  2. Then there exist numbers $i$ and $j$, $i\neq j$, such that $\sigma(i)=j$
  3. Since $n\geq 3$ there exists another number $k$.
  4. Let $\tau=( i k)\in S_n$ and find a contradiction.

Proof.

Seeking a contradiction, assume that the center $Z(S_n)$ is non-trivial.
Then there exists a non-identity element $\sigma \in Z(G)$.

Since $\sigma$ is a non-identity element, there exist numbers $i$ and $j$, $i\neq j$, such that $\sigma(i)=j$.

Now by assumption $n \geq 3$, there exists another number $k$ that is different from $i$ and $j$. Let us consider the transposition $\tau=(i k)\in S_n$.
Then we have
\begin{align*}
\tau \sigma (i)&=\tau (j)=j \\
\sigma \tau (i) &= \sigma (k) \neq j
\end{align*}
since $\sigma(i)=j$ and $\sigma$ is bijective.

Thus $\tau \sigma \neq \sigma \tau$ but this contradicts that $\sigma \in Z(S_n)$.
Therefore $Z(S_n)$, $n \geq 3$, must be trivial.


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