Stabilizer of an Element in Group Operations | Group Theory Homework

[hitmeoff]

Homework Statement

An operation of a group G on a set S is a function G X S \rightarrow S satisfying:

1. es = s \foralls \epsilon S
2. g(hs) = (gh)s \forallg,h \epsilon G, s\epsilon S

If s \epsilon S, show that the stabilizer of s, defined as the set:
{g \epsilon G | gs = s}
is a subgroup of G

Homework Equations

The Attempt at a Solution

Well from that definition it seems that g must be the identity element of G. Is a set consisting of just the identity not just a group? And since G is a group it includes the identity, thus the set g: {e} is a subgroup of G?

 

[lanedance]

i'm not sure about that...

my group theory isn't the best, but... i think the identity will surely be a part (from the defintion), but also potentially other elements as well

for example consider the group of all 2D rotations, I'm not 100% sure how to put it together, but imagine the set as a square, defined only by the location of its 4 vertices, only rotations of n*(pi/2) take the square into itself... i think the the stabiliser group in this case, then represents the rotational symmetry group of the square...

so i think the key is to show something along the lines of:
- the identity must be in the stabiliser group, and if f,h are in the stabiliser group, so is fh & h^{-1}

 

[lanedance]

once again expanded above

 

[HallsofIvy]

Homework Statement

An operation of a group G on a set S is a function G X S \rightarrow S satisfying:

1. es = s \foralls \epsilon S
2. g(hs) = (gh)s \forallg,h \epsilon G, s\epsilon S

If s \epsilon S, show that the stabilizer of s, defined as the set:
{g \epsilon G | gs = s}
is a subgroup of G

Homework Equations

The Attempt at a Solution

Well from that definition it seems that g must be the identity element of G. Is a set consisting of just the identity not just a group? And since G is a group it includes the identity, thus the set g: {e} is a subgroup of G?

No, the set of all g obviously contains e but there may be other members of G that "fix" s. Remember that s is NOT itself a member of G. You can only say that es= s because of (1) in your definition of the action of G on S.

(1) tells you that e is in this set and (2) tells you that the operation is associative. Now you need to prove that this set is closed under the operation: if gs= s and hs= s, then (gh)s= s. You also need to prove that if gs= s, then g^-1s= s.