Let G be a group acting on a set A and fix some a in A. Show that the following sets are subgroups of G.
(a) the kernel of the action
(b) stabilizer of a in G.
I am not sure what needs to be done.
an action of G on A is, by definition, a homomorphism , (Sym(A) = permutations of A, which may be an infinite set). often is written g.a, g(a), or ga.
some texts write this as:
an action is a mapping from GxA --> A (written ga) such that:
1) g(ha) = (gh)a (this is equivalent to saying ), for all g,h in G and a in A.
2) ea = a, for all a in A.
so part (a) is obvious from the first definition of group action. the second definition is what leads to the set girdav has given. it's fairly trivial to show that if ga = a for all a in A, and ha = a, for all a in A, then (h^-1)a = h^-1(ha) = (h^-1h)a = ea = a, so h^-1 is in the kernel if h is, and thus:
(gh^-1)a = g((h^-1)a) = ga = a, so we indeed have a subgroup.
the subgroup in (b) is often denoted Stab(a), or .
suppose h is in Stab(a), so ha = a. then (just as above) h^-1(a) = h^-1(ha) = (h^-1h)a = ea = a, so h^-1 stabilizes a whenever h does.
(this is just like with the kernel, except here we are only dealing with a particular a in A, rather than every a in A). in particular the kernel (often written Fix(A)), is a subgroup of Stab(a) for every a in A, since the kernel stabilizes everything).
so if g,h are in Stab(a), then (gh^-1)a = g(h^-1(a)) = ga = a, so Stab(a) is a subgroup.
(incidentally, group actions should have a familiar look to you. if our set is a vector space V, and our group is the multiplicative group F* of a field, then scalar multiplication defines an action of F* on V).