Conjugacy class of maximum order in $S_n$ Apparently it is well known that the conjugacy class of maximum order is the one with one cycle of length $n-1$ and one element fixed. Which is somewhat intuitive when you already know the answer since we want $1^{c_1}c_1!\cdot2^{c_2}c_2!\cdot\ldots\cdot n^{c_n}c_n!$ to be minimal.
However I wasn't able to find a formal proof anywhere on the internet. This question was asked here once and wasn't answered. I was thinking that one probably could make an injective and non surjective mapping from all other conjugacy classes but I'm pretty sure that is not an elementary approach and I have no idea how I would even begin to do that. I'd rather have some kind of combinatorial proof if possible.
Edit: I know how to find order of the specific conjugacy class in $S_n$. What I want to know is how would someone deduce that the one of the highest order is the conjugacy class consisting of a cycle of length $n-1$ and one fixed element.
 A: This is not hard to prove, although it's a bit tedious.
It might be easier to show that the $(n-1)$-cycle has the smallest centralizer, which has order $n-1$ when $n>2$, which we can do by induction on $n$. (When $n=2$, all centralizers have order $2$.)
Let $g \in S_n$. If $g$ is an $n$-cycle then its centralizer has order $n$ which is larger than $(n-1)$, so assume that $g$ is neither an $n$-cycle nor an $(n-1)$-cycle.
Then $g$ fixes two disjoint subsets $A$ and $B$ of $A \cup B =\{1,2,\ldots,n\}$ with $2 \le a  \le b \le n-2$, where $a=|A|$ and $b=|B|$, and $$|C_{S_n}(g)| \ge |C_{{\rm Sym}(A)}(g|_A)||C_{{\rm Sym}(B)}(g|_B)|\ge (a-1)(b-1)$$ by inductive assumption, but when $a=2$ we can replace the right hand side by $2(b-1)$.
Now if $a=2$ and $b \ge 4$, then $2(b-1) > n-1 = a+b-1$, whereas if $a \ge 4$, or $a=3$ and $b \ge 5$, then $(a-1)(b-1) > a+b-1$, so this leaves only a few small cases like $a=b=3$ to check individually.
A: Consider the denominator $\prod_jj^{c_j}c_j!$ you mentioned, which needs to be minimized. Since this is larger for an $n$-cycle than for an $(n-1)$-cycle (for $n\ge3$), the maximal conjugacy class isn’t the class of $n$-cycles. Thus there is a largest cycle of length $k$, and at least one other cycle of length $l$. Combining these two cycles into a single cycle creates a cycle of a length $k+l$ that hadn’t occurred before. Thus, in the resulting permutation $c_{k+l}=1$, and $c_k$ and $c_l$ haven’t increased, so the product $\prod_jc_j!$ hasn’t increased. The product $\prod_jj^{c_j}$ has changed by a factor $\frac{k+l}{kl}=\frac1k+\frac1l$, which is $\le1$ unless one of $k$ and $l$ is $1$. Thus the denominator can always be decreased unless the only cycles except the largest one are $1$-cycles. Merging one of the $1$-cycles into the largest cycle (of length $k$) changes the denominator by a factor $\frac{k+1}{kc_1}$, which is $\gt1$ for $c_1=1$ but $\le1$ otherwise. Thus, the permutations in the maximal congugacy class have exactly one $1$-cycle in addition to the largest cycle of length $k=n-1$.
