Question about a proof showing that the center of $S_n$ is trivial How do I need to modify this in order for it to be correct?
The center of $S_n$ (for $n\geq$ 3) is the trivial identity. Proof: Assume the center of $S_n$ is $C = \{ id , \tau \}$ where $ \tau \in S_n$ and $\tau \neq \ id$. Then for some $n$ the factor group $S_n\backslash C$ is abelian and solvable, a contradiction. 
 A: This formula should be helpful.
Suppose $\sigma,\tau\in S_n$, with $\tau=(a_1\ldots a_m)$ an $m$-cycle.  Then:
$$\sigma\tau\sigma^{-1}=(\sigma(a_1)\ldots\sigma(a_m))$$
A: "Let $S$ be a set having more than two elements , then the center of $A(S)$ , the permutation group of $S$ , is trivial" . 
Let $ id \ne \sigma \in A(S) $ , then $\exists a \in S $ such that $b:=\sigma(a)\ne a$ . As $S$ has more 
than two elements , 
$\exists c \in S $ such that $c \ne a , c \ne b$ . Now consider the transposition 
$\tau:=(b,c) \in A(S)$ . Now 
$\tau(b)=c , \tau(c)=b$ and since $a \ne b$ and $a \ne c$ , we get $\tau (a)=a$ . We see 
that $(\sigma \circ \tau) (a)=\sigma(\tau(a))=\sigma(a)=b $ and $(\tau \circ \sigma)(a)=\tau(\sigma(a))=\tau(b)=c$ ; since $c \ne b$ , we 
conclude $(\sigma \circ \tau) (a) \ne (\tau \circ \sigma)(a)$ i.e. $\sigma \circ \tau \ne \tau \circ \sigma$ , thus $\sigma \notin Z(A(S))$ , hence 
$id \ne \sigma \in A(S) \implies \sigma \notin Z(A(S))$
A: Let me use the general idea you are trying to use to prove the result. This is very different from the idea which Jared is proposing. My proof uses the fact that $A_n$ is simple, which means that it only works for $n\geq 5$. It doesn't use anything you couldn't have thought up yourself, and it is interesting and different enough from Jared's method for the $n\geq 5$ not to really matter. Just brute-force $n=3$ and $n=4$ (which we do at the bottom).
Suppose $1\neq g\in Z(S_n)$, then $g\not\in A_n$ but $g^2\in Z(A_n)$ (why?). As $A_n$ is simple, this means that $g^2$ is trivial. Thus, every element of the centre must have order two. As $g\not\in A_n$, we can re-labelling the underlying set (this is an automorphism of $S_n$) to write $g=(1,2)(3,4)\ldots (m, m+1)$ where $m$ is odd. Note that if $m>1$ we can re-label the underlying set to get that $(1,3)(2,4)\ldots (m, m+1)\in Z(G)$ also (why?), while if $m=1$ then by the same trick we get that $(3, 4)\in Z(S_n)$. Therefore, $|Z(G)|>2$.
Consider $N=\langle g\rangle$. Note that $N$ is a normal subgroup of $S_n$ of order two, with trivial intersection with $A_n$. Thus, $S_n\cong A_5\times C_2$ (this uses the fact that $S_n=A_nN$. Can you see why this holds?). This means that $|Z(S_n)|=2$ (why?). We therefore have a contradiction.
EDIT: To make the proof complete, we'll cover $n=3$ and $n=4$.
For the $n=3$ case, note that $(1, 2)$ does not commute with $(1, 2, 3)$. This is sufficient (why?). Alternatively, one can use the fact that if $G/Z(G)$ is cyclic then $G$ is abelian, which is a classic undergraduate question so I'll leave it for you to prove! This fact means that any group of order $pq$, $p$ and $q$ primes, is either abelian or has trivial centre.
For $n=4$, we'll prove that $A_4$ has trivial centre. This is enough, by the working above (where we only use the fact that $A_n$ has trivial centre - the whole "simple" bit is really overkill!). To see that $A_4$ has trivial centre, note that it has order twelve and its elements have the form $(a, b)(c, d)$ or $(a, b, c)$. Note that if $(a, b, c)\in Z(G)$ then every element of this form is in $Z(G)$, and similarly for $(a,b)(c, d)$. This is because permuting the elements of the base set $\{1, 2, 3, 4\}$ is an automorphism of $A_4$ (just as it was of $S_n$ earlier), and because $Z(A_4)$ is characteristic. Therefore, we have four choices for $Z(A_4)$. Either it is trivial, or is the whole of $A_4$, or $(1,2,3)\in Z(A_4)$ but $(1, 2)(3, 4)\not\in Z(A_4)$ or $(1,2)(3,4)\in Z(A_4)$ but $(1, 2, 3)\not\in Z(A_4)$. Therefore, either $Z(A_4)$ is trivial or $(1, 2, 3)$ and $(1, 2)(3, 4)$ commute. These do not commute, and so $Z(A_4)$ is trivial, as required.
