find the centralizer Z(a) of the element a=(134) in S5

Printable View

- Feb 25th 2009, 05:05 AMwonjujincentralizer
find the centralizer Z(a) of the element a=(134) in S5

- Feb 26th 2009, 02:58 PMThePerfectHacker
Let $\displaystyle [(134)]$ be the conjugacy class of $\displaystyle (134)$ and let $\displaystyle C(134)$ be the centralizer of that element. Then, the number of elements of $\displaystyle [(134)]$ is equal to the index of $\displaystyle C(134)$ in $\displaystyle S_5$. Now the number of conjugacy classes of $\displaystyle (134)$ is equal to $\displaystyle 2! \cdot {5\choose 3} = 20$. Therefore, the index of $\displaystyle C(134)$ in $\displaystyle G$ is equal to $\displaystyle \tfrac{5!}{20} = 6$. This tells us that $\displaystyle C(134)$ has six elements. Let us see if we can figure them out through reasoning as opposed to mindless computation of all 120 elements in $\displaystyle S_5$. First, certainly $\displaystyle \left< (134)\right> \subset C(134)$ and so $\displaystyle \text{id},(134),(134)^2$ are all elements of the centralizer. Second, $\displaystyle (25)$ is a disjoint cycle from $\displaystyle (134)$ therefore $\displaystyle (25)(134) = (134)(25)$. Thus, we see that $\displaystyle \text{id},(134),(134)^2,(25),(25)(134),(25)(134)^2$ all commute with $\displaystyle (134)$. We found six elements, therefore, those elements mentioned above is the centralizer.