Prove that the symmetric group $S_n$, $n \geq 3$, has trivial center. [duplicate]

I am trying to prove this:

Let $\sigma$ be a non-identity element of $S_{n}$. If $n \geq 3$ show that $\exists \gamma \in S_{n}$ such that $\sigma\gamma \neq \gamma\sigma$.

Hint: Let $\sigma*k=1$ where $k \neq 1$. Let $m$ not be an element of $\{k,1\}$. Let $\gamma=(k,m)$

Here's the answer from the textbook:

Let $\sigma*k=1$ for some $k \neq 1$. Then as $m\geq 3$, choose an $m$ not an element of $\{k,1\}$. Now let $\gamma=(k,m)$. This gives $\gamma*\sigma*k=\gamma*(1)=1$ but $\sigma*\gamma*k=\sigma*m \neq 1$, since if $\sigma*m=1=\sigma*k$, then $m=k$, as $\sigma$ is one-to-one contrary to the assumption.

I guess I'm just not understanding the steps to their proof. Can anybody make sense of this to someone new to proofs? I mostly don't understand how $\sigma*\gamma*k=\sigma*m=1$

Solution 1:

Here is the proof written out hopefully more clearly and with slightly more details.

Since $\sigma\neq \epsilon$, without loss of generality we may assume $\sigma(1)\neq 1$. Then there exists $k\neq 1$ with $\sigma(k)=1$. Choose $m\in\{1,\ldots,n\}\setminus\{1,k\}$, and set $\gamma=(k,m)$ (a transposition). Then $\gamma\circ \sigma(k)=\gamma(1)=1$. On the other hand, $\sigma\circ\gamma(k)=\sigma(m)$. As $\sigma$ is injective, $\sigma(k)=1$ and $m\neq k$, we know that $\sigma(m)\neq 1$. Hence $\gamma\circ\sigma\neq\sigma\circ\gamma$ (as these two maps take different values at $k$).

Solution 2:

I think this is easiest to understand if you use the fact that if $(a_1... a_k)$ is a cycle of $\sigma$ then $(\gamma a_1 ... \gamma a_k)$ is a cycle of $\gamma \sigma \gamma^{-1}$.

$\sigma$ being in the center of $S_n$ is equivalent to the statement that $\sigma = \gamma \sigma \gamma^{-1}$ for all $\gamma \in S_n$. So to show $\sigma \neq 1$ is not in the center of $S_n$, you just have to find some $\gamma$ and some $k$-cycle $(a_1... a_k)$ of $\sigma$, where $k \geq 2$, for which $(\gamma a_1 ... \gamma a_k)$ is not a cycle of $\sigma$.

This is easy to do; if $(a_1... a_k)$ is a cycle of $\sigma$ you can for example let $\gamma a_1 = a_1$ and let $\gamma a_2$ be something outside the cycle, or in the event that there's exactly one cycle (so that $k = n$) just let $\gamma a_2 = a_n$.

Solution 3:

It is enough to show that $S_3$ is not abelian, since $S_n$ always has a subgroup isomorphic to $S_3$ for $n\ge 3$. An abelian group could not have a non-abelian subgroup. And for $S_3$ we compute $$ (12)(13)=(132)\neq (123)=(13)(12). $$

Solution 4:

Consider two permutations

$\sigma =(a$ $ b)$

$\gamma =(b$ $a$ $c$ $\cdots \cdots$)

Note gamma can be of any length $(n\geq 3)$.

$\sigma \circ \gamma (a) = c$

$\gamma \circ \sigma (a) = a$

Thus $\sigma \circ \gamma \neq \gamma \circ \sigma$

Thus no $two-cycle$ commutes with all the elements. Also, no $n-cycle$ commutes with all the elements.