# Operations on a group

1. Nov 6, 2009

### hitmeoff

1. The problem statement, all variables and given/known data

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

1. es = s $$\forall$$s $$\epsilon$$ S
2. g(hs) = (gh)s $$\forall$$g,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

2. Relevant equations

3. 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?

2. Nov 6, 2009

### lanedance

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}

Last edited: Nov 6, 2009
3. Nov 6, 2009

### lanedance

once again expanded above

4. Nov 6, 2009

### HallsofIvy

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.