I tried proving that for all $n \in \mathbb{N}_{>2}$ the centre $Z(S_{n})$ of the symmetric group $S_{n}$ is trivial. I already have a proof, but I'm not actually sure whether it's correct. Would someone like to proofread it?
I tried proving it by looking at conjugation and order. Let $\sigma \in Z(S_{n})$ and $\tau \in S_{n}$ be randomly given. Let $s$ be the order of $\sigma$, $s^{-1}$ the order of $\sigma^{-1}$ and $t$ the order of $\tau $.
We know that $\tau \in S_{n}$ , so $\sigma \cdot \tau \cdot \sigma^{-1} \in S_{n}$. We know that $\sigma$ commutes with everything, so $\sigma \cdot \tau \cdot \sigma^{-1} = \tau$ So $\tau$ and $\sigma \cdot \tau \cdot \sigma^{-1}$ both have order $t$.
We also know that $\sigma \cdot \tau \cdot \sigma^{-1}$ its order is $lcm(s, lcm(t, s^{-1}))$. But the least common multiple is both associative and commutative, so $\sigma \cdot \tau \cdot \sigma^{-1}$ its order is also $lcm(lcm(s,s^{-1}), t)$. But we know that $lcm(s,s^{-1}) = 1$, considering $\sigma \cdot \sigma^{-1} = id$. So therefore both $s$ and $s^{-1}$ must be equal to $1$, but that also means that $\sigma = id$. But then all elements in $Z(S_{n})$ must be the identity, which was what we wanted to proof.
My biggest problem with this proof is the part where I state that because the $lcm$ is commutative and associative, $lcm(s,s^{-1}) = 1$, is this like a legit move? I don't know why but this just rubs me the wrong way
Unfortunately, there are a number of problems with your "proof". First, $s^{-1}=s$ (and the notation is very regrettable), so your statement that $lcm(s,s^{-1})=1$ is not legitimate. However, the most serious mistake is that you have not used the fact that $n\geq 3$.
Try the following:
Suppose $\sigma\in Z(S_n)$ is not the identity. Then $\sigma(i)=j\neq i$ for some $i,j\in\{1,\ldots,n\}$. Now, since $n\geq 3$, there exists $k\in \{1,\ldots,n\}\backslash\{i,j\}$.
Claim: There exists $\tau\in S_n$ such that $\tau(j)=k$, but $\tau(i)=i$ (can you think of one?).
Given the claim, compute $\sigma\tau(i)$ and $\tau\sigma(i)$. Are they equal? If not, what can you conclude?