Find the centralizer in $S_7$ of $(123)(4567)$.

Question

I am struggling to understand what the centralizer in a permutation of order $7$ means. "The centralizer consists of all elements that commute with $(123)(4567)$" but.. is there a more rigid mechanical way of solving this problem, akin to calculus?

Answer 1

Hint: Two elements $\theta$ and $\sigma= (1,2,3)(4,5,6,7)$ commute iff $\theta \sigma \theta^{-1}=\sigma = (1,2,3)(4,5,6,7)$. There is a result that says $\theta \sigma \theta^{-1}=(\theta(1),\theta(2),\theta(3))\cdot (\theta(4),\theta(5),\theta(6),\theta(7))=(1,2,3)(4,5,6,7)$.

The problem is therefore reduced to calculating a $\theta$ such that $(\theta(1),\theta(2),\theta(3))=(1,2,3)=(3,1,2)=(2,3,1)$ and $(\theta(4),\theta(5),\theta(6),\theta(7))=(4,5,6,7)=\cdots$. Can you take it from here?

Answer 2

Note that

$$C_{S_7}[(123)(4567)] = \{\sigma \in S_7: \sigma(123)(4567) = (123)(4567)\sigma\}$$

Clearly, $(123)(4567)$ commutes with itself and any power of itself; so $\langle (123)(4567) \rangle \leq C_{S_7}(\sigma)$. In general, if a permutation $\sigma \in S_n$ is written in its cycle decomposition $\sigma= \underbrace{\tau_1 \tau_2 \dots \tau_k}_{\mathrm{disjoint}}$, then since disjoint cycles commute, if you pick any $\tau_i$ from the above decomposition, we're guaranteed

$$\tau_i \sigma = \sigma \tau_i$$

i.e. we get $\tau_i \in C_{S_n}(\sigma)$ for every $i$ (and every power of such a $\tau_i$). Then it follows that

$$\langle \tau_1, \tau_2, \dots, \tau_k \rangle \leq C_{S_n}(\sigma)$$

This should cover most of the elements you're looking for. Can you rule out whether this misses any other elements?