Stabiliser of $(12345)\in S_5$ for conjugation.

Question

Let $\sigma= (12345)\in S_5$. I read that the Stabiliser of $\sigma$ under conjugation is the group generated by $\sigma$.

Can someone explain how to prove this?

I know that $\tau \sigma \tau^{-1}= (\tau(1)\tau(2)\tau(3)\tau(4)\tau(5))$ and want this to be $(12345)$. But how does this help?

Answer 1

Note that cycle notation for a permutation is unique up to a cyclic shift of its entries in each cycle and permutation of cycles.

So in order for $(\tau(1)\tau(2)\tau(3)\tau(4)\tau(5))$ to be $(1 2 3 4 5)$, given $\tau(1)$, you know $\tau$. Thus the stabilizer of $\sigma$ under conjugation (also known as the centralizer of $\sigma$ in $S_5$) contains at most $5$ elements and definitely contains the subgroup generated by $\sigma$ which is a cyclic group of order $5$.

Answer 2

By the orbit stabiliser theorem, the size of the orbit is the index of the stabiliser.

But, by a well-known fact, and the equation you wrote down, the orbit consists of all of the $5$-cycles, of which there's $4!.$

Thus the stabiliser has order $5!/4!=5.$

But clearly any power of $(12345)$ is in the stabiliser.