Find the conjugacy class in $S_n$ of $(12...n)$. What is the order of the centalizer of $(12...n)$ in $S_n$?

Question

Ok so first thing I don't understand is why centralizer of $(12...n)$ has $n$ elements. Shouldn't equation $(\phi(1)\phi(2)...\phi(n))=(12...n)$ have more solutions? And is my reasoning here correct: Conjugacy class of $(12...n)$ contains every permutation consisting of $n$-cycle ?

Answer 1

One way to approach things is to find the conjugacy class first (you're right about that part, by the way). Then consider the action of $S_n$ on itself by conjugation. The centralizer of an element is its stabilizer under this action, and you can find the order of the stabilizer by applying the orbit-stabilizer theorem.

Specifically, the size of the orbit of $(12...n)$ is equal to the size of its conjugacy class, which is $(n-1)!$.

[This is the number of $n$-cycles in $S_n$: there are $n!$ possible orderings of the terms $1$, $2$, ..., $n$, but each permutation is represented by $n$ distinct orderings (namely, the orderings each starting with a different number at the beginning but proceeding in the same order, e.g. $1$, $2$, ..., $n$ and $2$, $3$, ..., $n$, $1$ represent the same permutation). So there are $n!/n = (n-1)!$ distinct $n$-cycles.]

Then, by the orbit-stabilizer theorem, the stabilizer of $(12...n)$ contains $|S_n|/(n-1)! = n!/(n-1)! = n$ elements.

Using your formulation of the question ("Why doesn't $(\varphi (1)\varphi (2)...\varphi (n))=(12...n)$ have more than $n$ solutions?"), consider the reasoning we used above to count the number of $n$-cycles. This equation will hold precisely when $\varphi$ reorders $1$, $2$, ..., $n$ in such a way that $(\varphi (1)\varphi (2)...\varphi (n))$ represents the same permutation as $(12...n)$, which means that $\varphi$ must preserve the ordering of $(12...n)$ but simply begin the cycle with a different element, e.g. $(23...n1)$. There are only $n$ ways to do this.