I tried to answer the following question:
Why does the fact that the orders of the elements of $A_4$ are $1,2$ and $3$ imply that $|Z(A_4)|=1$?
My answer: Two cycles commute if and only if they are disjoint. But no combination of $2$-cycles and $3$-cycles in $A_4$ can ever be disjoint since there are only $4$ numbers. Hence the centre must be trivial.
Then I looked at the answer:
If there was $x$ in $Z(A_4)$ of order $2$ then a product with a $3$-cycle would have order $6$.
If there was $x$ in $Z(A_4)$ of order $3$ then a product with a $2$-cycle would have order $6$.
I have two questions about this question:
(1) Is my proof correct?
and:
(2) Is it a general principle that if the goal is to show the centre of $S_n$ is trivial one assumes by contradiction it wasn't and then produces an element with order too large to be in $S_n$?