Center of Symmetric Group is Trivial

Theorem
The center of the Symmetric Group of order 3 or greater is trivial.

Proof
Clearly the identity lies in the center.

Assume $$ g $$ is not the identity, so that $$ g(a) = b \neq a $$ for some $$ a $$.

As the order is greater than $$ 2 $$, we may choose $$ c \neq b,g(b) $$ and set $$ h = (b\ c) $$.

Then, $$ hg(b) = g(b) \neq g(c) = gh(b) $$ (recall that permutations are one-to-one.)

This proves that $$ g $$ cannot be in the center and we're done.