If elements of $S_n$ commute then they are disjoint?

Question

Why does the fact that the orders of the elements of $A_4$ are $1, 2,$ and $3$ imply that $|Z(A_4)|=1?$

In solution provided it say that If there exits $x \in Z(A_4)$ and order of $x$ is 2 or 3 then by composing $x$ with other element of order $3$ or $2$ is respectively will give us an element of order of $6$ (since $x$ commute with all its order is LCMs of order of elements)

I am confused with this reason because according to theorem criteria of using LCMs is used when the cycles are disjoint and if cycles are disjoint then they commute. So my question is that is the converse is true that if elements of $S_n$ commute then they are disjoint?

Answer 1

No, $(1,2)$ commutes with $(1,2)$.

Answer 2

This proof does not appeal to any particular theorem about orders of elements in $S_n$. Instead, it uses the much more general fact that if $G$ is a group, and $g, h \in G$ with $\operatorname{ord}(g) = n$, $\operatorname{ord}(h) = m$, where $n$ and $m$ are coprime, and $g$ and $h$ commute, then $\operatorname{ord}(gh) = nm$.

This is because $(gh)^k = g^k h^k$. This is fairly obvious from the fact that they commute, so you can just move all the $k$ $g$s to one side, but if you really want you can prove this by induction.

Then clearly if $(gh)^k = e$, we have $g^k h^k = e$. Taking $n$th powers, we get $h^{nk} = e$. Then we must have that $m$ divides $nk$, but $m$ is coprime to $n$, so $m$ divides $k$. Similarly, by taking $m$th powers, we get that $n$ divides $k$. So $nm$ divides any such $k$. Furthermore, clearly if both $n$ and $m$ divide $k$, then $(gh)^k = e$. So by definition, $\operatorname{ord}(gh) = nm$, as this is the least common multiple of $n$ and $m$.

This may work under some weaker assumptions, but in your case I think the proof is just assuming that it's obvious that the product of commuting elements of orders $2$ and $3$ respectively has order $6$. (Remember that $x \in Z(A_4)$, so it commutes with everything)