prove H_T less than or equal to S_n

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October 13th 2008, 12:00 PM
mandy123

prove H_T less than or equal to S_n

Let T=[(a_1, a_2,...,a_t) C {1,2,...n}] and H_T={beta is a member of S_n|beta(a_i)=a_i, for all a_i which is a member of T}. H_t is called the stabilizer of T in S_n, stab(T). Prove that H_T < S_n. What is the order of H_T
October 13th 2008, 12:16 PM
Jhevon

do you understand what H_T does?

Quote:

Originally Posted by mandy123

Let T=[(a_1, a_2,...,a_t) C {1,2,...n}] and H_T={beta is a member of S_n|beta(a_i)=a_i, for all a_i which is a member of T}. H_t is called the stabilizer of T in S_n, stab(T). Prove that H_T < S_n.

it suffices to show that H_T is a subgroup of S_n. or do you know that and are wondering how to do it?

you must show that H_T is nonempty. (consider the identity function)

then show that it is closed under the composition of functions

then show that each element has an inverse

which of these do you have problems with?

Quote:

What is the order of H_T

H_T leaves the elements in T fixed and permutes the other elements. how many ways can we permute the remaining elements?
October 13th 2008, 05:41 PM
mandy123

ok to say it is nonempty:
beta(a_i)=a_i, so a_i^2=a_i So H is nonempty?

to show closure:
x,y is a member of H_t
beta(x)=a_i and beta(y)=a_i

Inverse: Dont know how to do this part

Did I get the other 2 right?
I dont understand this stuff.(Headbang)
November 16th 2008, 02:59 PM
Jhevon

Quote:

Originally Posted by mandy123

ok to say it is nonempty:
beta(a_i)=a_i, so a_i^2=a_i So H is nonempty?

to show closure:
x,y is a member of H_t
beta(x)=a_i and beta(y)=a_i

Inverse: Dont know how to do this part

Did I get the other 2 right?
I dont understand this stuff.(Headbang)

note that the identity function is in $H_T$ so that it is not empty

to show closure, we need to show that if $\alpha (a_i) ,~\beta (a_i) \in H_T$ then (without loss of generality) $\beta \circ \alpha \in H_T$

so assume $\alpha , ~\beta \in H_T$ . then $\alpha (a_i) = a_i$ and $\beta (a_i) = a_i$

thus $( \beta \circ \alpha ) (a_i) = \beta ( \alpha (a_i)) = \beta (a_i) = a_i$

so that $\beta \circ \alpha \in H_T$ since $( \beta \circ \alpha )(a_i) = a_i$

For inverses, note that if $\alpha \in H_T$ then $\alpha ^{-1} \in S_n$ works as the inverse of $\alpha$ , and it is also in $H_T$ , since if we assume $\alpha \in H_T$ , then

$\alpha (a_i) = a_i$

$\Rightarrow \alpha ^{-1} (\alpha (a_i)) = \alpha ^{-1} (a_i)$

$\Rightarrow a_i = \alpha ^{-1}(a _ i)$

so that $\alpha ^{-1} \in H_T$

thus, $H_T \le S_n$