TITLE: Conjugacy class of maximum order in $S_n$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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 $k=n-1$.