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
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
do you understand what H_T does?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?
H_T leaves the elements in T fixed and permutes the other elements. how many ways can we permute the remaining elements?What is the order of H_T
note that the identity function is in $\displaystyle H_T$ so that it is not empty
to show closure, we need to show that if $\displaystyle \alpha (a_i) ,~\beta (a_i) \in H_T$ then (without loss of generality) $\displaystyle \beta \circ \alpha \in H_T$
so assume $\displaystyle \alpha , ~\beta \in H_T$. then $\displaystyle \alpha (a_i) = a_i$ and $\displaystyle \beta (a_i) = a_i$
thus $\displaystyle ( \beta \circ \alpha ) (a_i) = \beta ( \alpha (a_i)) = \beta (a_i) = a_i$
so that $\displaystyle \beta \circ \alpha \in H_T$ since $\displaystyle ( \beta \circ \alpha )(a_i) = a_i$
For inverses, note that if $\displaystyle \alpha \in H_T$ then $\displaystyle \alpha ^{-1} \in S_n$ works as the inverse of $\displaystyle \alpha$, and it is also in $\displaystyle H_T$, since if we assume $\displaystyle \alpha \in H_T$, then
$\displaystyle \alpha (a_i) = a_i$
$\displaystyle \Rightarrow \alpha ^{-1} (\alpha (a_i)) = \alpha ^{-1} (a_i)$
$\displaystyle \Rightarrow a_i = \alpha ^{-1}(a _ i)$
so that $\displaystyle \alpha ^{-1} \in H_T$
thus, $\displaystyle H_T \le S_n$