# elements of symmetric groups and centralizers

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• Apr 10th 2010, 01:48 PM
kimberu
elements of symmetric groups and centralizers
Let P = (1, 2, 3, .... n) in $S_n$.
a. Show there are (n-1)! distinct n-cycles in $S_n$.
b. How many conjugates does P have in $S_n$?
c. Let C(P)=N(P) be the centralizer of P in $S_n$. Using (b), find |C(P)|.
d. Find the subgroup C(P).

my thoughts:
a -- this seems really obvious, but I never actually learned how to show it! I have no ideas.
b. by part (a), wouldn't this be simply (n-1)!, since that's the property of symmetric groups?
c & d, I really don't know what to do. Do I use the class equation somehow, or another theorem?

Thanks so much for any help!
• Apr 10th 2010, 09:54 PM
aliceinwonderland
Quote:

Originally Posted by kimberu
Let P = (1, 2, 3, .... n) in $S_n$.
a. Show there are (n-1)! distinct n-cycles in $S_n$.
b. How many conjugates does P have in $S_n$?
c. Let C(P)=N(P) be the centralizer of P in $S_n$. Using (b), find |C(P)|.
d. Find the subgroup C(P).

my thoughts:
a -- this seems really obvious, but I never actually learned how to show it! I have no ideas.
b. by part (a), wouldn't this be simply (n-1)!, since that's the property of symmetric groups?
c & d, I really don't know what to do. Do I use the class equation somehow, or another theorem?

Thanks so much for any help!

For (a), observe that (1 2 3 ... n )= (2 3 .. n 1) = (3, 4, .. n 1 2 ). To find the number of distinct n-cycles, you need to fix the first element in the n-cycle, let's say 1, and consider the permutation of the remaining elements. So there are (n-1)! distinct n-cycles in S_n.

For (b), a conjugacy class in S_n has the same cycle type. So there are (n-1)! memebers in the conjugacy class of n-cycle in S_n.

(c). The centralizer of (1 2 3 .. n) in S_n is simply a group generated by (1 2 3 .. n), which is <(1 2 3 .. n)>. If this is not immediate for you, cosider the permutation with two line notation.

The members in the group <(1 2 3...n)> have a pattern in the two line notation. Try some members in the above group with two line notation and observe how they commute with (1 2 3 .. n).

However, if you need to find the centralizer of (1 2 3 .. m) in S_n with m<n, the above needs a little bit of modification.

For instance, the centralizer of (1 2 3 4 5 ) in S_8 is the group <(1 2 3 4 5), (6 8), (7 8)>.
• Apr 11th 2010, 12:26 AM
kimberu
Quote:

Originally Posted by aliceinwonderland

(c). The centralizer of (1 2 3 .. n) in S_n is simply a group generated by (1 2 3 .. n), which is <(1 2 3 .. n)>. If this is not immediate for you, cosider the permutation with two line notation.

The members in the group <(1 2 3...n)> have a pattern in the two line notation. Try some members in the above group with two line notation and observe how they commute with (1 2 3 .. n).

Thanks a lot - I understand parts a&b, but I'm a little shaky on the centralizer. I don't think I know which unique pattern you mean in the two-line notation. (Also, is the order of <(1 2 3 .. n)> also (n-1)!, that is, is this group the same as the conjugacy class? I'm not sure how to find that either.)
• Apr 11th 2010, 02:15 AM
aliceinwonderland
Quote:

Originally Posted by kimberu
Thanks a lot - I understand parts a&b, but I'm a little shaky on the centralizer. I don't think I know which unique pattern you mean in the two-line notation. (Also, is the order of <(1 2 3 .. n)> also (n-1)!, that is, is this group the same as the conjugacy class? I'm not sure how to find that either.)

The order of <(1 2 3 ... n)> is n. It is the cyclic group generated by (1 2 3 .. n) in S_n. So it is an abelian group.

For instance, if n=4 and let $\tau = (1 2 3 4 )$, then the cyclic group generated by $\tau$ is $\{1, \tau, \tau^2, \tau^3\}$. I encourage you to verify this with the two line notation. All elements in this group commute with $\tau$, which implies $\rho\tau\rho^{-1}=\tau$ where $\rho \in \{1, \tau, \tau^2, \tau^3\}$.