Proposition For any 5 € S, the set stab(s) is a subgroup of G. Proof (outline) To show stab(s) is a group, we need to show three things: (i) Identity. That is, s.6(1)=s. (ii) Inverses. That is, if s.4(g)=s, then s(9¯¹) = 5. (iii) Closure. That is, if s.a(g) =s and 5.0(n) =s, then s.d(gh) = s. Alternatively, it suffices to show that if s.(a) =s and s.4(h)=s, then s.0(gh¹) = 5. The stabilizer subgroup Here is another example of an action, this time of G = Do Let's be the highlighted hexagon, and H-stab(s). The stabilizer of s in G is stab(s) {g G|s.(g) = s}. The stabilizer subgroup As we've seen, elements in the same orbit can have different stabilizers Proposition (HW exercise) Set elements in the same orbit have conjugate stabilizers: stab(s. (g)) = gstab(s)g. for all g€ G and s E S. Problem: Let G GL, (F) be the general linear group of n x n invertible matrices over a field F, and let EF" be a non-zero vector. Consider the group action of G on F" by matrix multiplication. • Prove that the stabilizer of tv in GL, (F) is isomorphic to GL-1(F). Calculate the order of the stabilizer if F = F, a finite field with g elements. Here is the action graph of the group D₁ = (r. f): Hr³ Hr2 HM Hr (14) In other words, if x stabilizes s, then glxg stabilizes s.$(g). Here are several ways to visualize what this means and why. 0(5) Hr5 (f) (124) (x) 5 (9) (91x9) +5' (0) (r) $(r) (fr) = (³) In the beginning of this course, we picked a configuration to be the "solved state," and this gave us a bijection between configurations and actions (group elements).