Centralizer in Symmetric Group of 7 elements.

Question

I'm currently working on trying to find the centralizer of the element $a=(123)(4567)$ in $S_7$ and my guess is that it is the group generated by a. The second inclusion is clear, but I'm have trouble showing that the centralizer is contained in the group generates by a. Here is my idea:

Given $b$ in the centralizer, it must be that $b^{-1}ab=a$, and so if $b^{-1}ab=c_1c_2$, where $c_1$ is a three cycle and $c_2$ is a four cycle, then $$ c_1=(b(1),b(2),b(3))=(123)=(231)=(312) $$ and $$ c_2=(b(4),b(5),b(6),b(7))=(4567)=(5674)=(6745)=(7456). $$

I'm not sure what to do from here. Any suggestions would he appreciated! $$

Answer 1

Notice that $bab^{-1}=(b(1),b(2),b(3))(b(4),b(5),b(6),b(7))$.

Therefore if $b$ centralizes $a$ we must have that $b$ permutes $\{1,2,3\}$ and $\{4,5,6,7\}$ among each other. And we must also have that $b$ acts on $\{1,2,3\}$ as a power of the cycle $(1,2,3)$ and acts on $\{4,5,6,7\}$ as a power of the cycle $(4,5,6,7)$. Every such permutation is a power of $a$.

We can generalize a bit.

Let $a$ be a permutation such that all of its $k$ cycle's lengths are the distinct. Then the centralizer of $a$ is the group generated by $c_1,c_2,c_3\dots c_k$. Where $c_j$ is the $j$'th cycle. And notice that if additionally all of the cycle lengths are pairwise coprime, this coincides with the subgroup generated by $a$.

Answer 2

You can continue as follows: if you write that $(b(1),b(2),b(3))=(123)$ this gives you $b(1)=1, b(2)=2$, and $b(3) = 3$ or $b = id$; when you write $(b(1),b(2),b(3))=(231)$ note that $b = (2,3,1)$ and so on to obtain $ (), (4,5,6,7), (4,6)(5,7), (4,7,6,5), (1,2,3), (1,2,3)(4,5,6,7), (1,2,3)(4,6)(5,7), (1,2,3)(4,7,6,5), (1,3,2), (1,3,2)(4,5,6,7), (1,3,2)(4,6)(5,7), (1,3,2)(4,7,6,5) $