Proposition For any s€ 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, 5.4(1) = s. (i) Inverses. That is if s.o(g)-s, then s.(g¹)=s. (iii) Closure. That is, if s.p(g) =s and s.(n) =s, then s.(gh) = s. Alternatively, it suffices to show that if s.(g) =s and s.d(h) = s. then s.0(gh¹) = 5. The stabilizer subgroup Here is another example of an action, this time of G = De- Let's be the highlighted hexagon, and Hstab(s). Hr² Hr (121) (148) Let G be a finite group acting on a set X, and let z X be an element of the set. The stabilizer of , denoted Stabc(z), is defined as the set of all elements in G that fix 2 under the group action, The stabilizer of s in G is i.e., stab(s) {gG|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 gЄ G and s € S. In other words, if x stabilizes s, then g¹xg stabilizes s. (g). Here are several ways to visualize what this means and why. Prove the following: Stabc() {gGg-2=2} The stabilizer of z is a subgroup of G. If G acts transitively on X, then the size of the orbit of a is given by G-z]=[G: Stab()], where [G: Staba (z)] denotes the index of the stabilizer subgroup in G. Here is the action graph of the group D₁ = {r, f): 2 Hr³ H+4 labeled by destinations Hr Н labeled by stabilizers 0(x) 5 $(1) (9) (9) $(r) (r) (fr)=(1) (124) In other words, if x is a loop from s, and ss, then g¹xg is a loop from s'. In the beginning of this course, we picked a configuration to be the "solved state," and this gave us a bijection between configurations and actions (group elements).
Proposition For any s€ S, the set stab(s) is a subgroup of G. Proof (outline) To show stab(s) is a group we need to show three things: (i) Identity. That is, 5.4(1) = s. (i) Inverses. That is if s.o(g)-s, then s.(g¹)=s. (iii) Closure. That is, if s.p(g) =s and s.(n) =s, then s.(gh) = s. Alternatively, it suffices to show that if s.(g) =s and s.d(h) = s. then s.0(gh¹) = 5. The stabilizer subgroup Here is another example of an action, this time of G = De- Let's be the highlighted hexagon, and Hstab(s). Hr² Hr (121) (148) Let G be a finite group acting on a set X, and let z X be an element of the set. The stabilizer of , denoted Stabc(z), is defined as the set of all elements in G that fix 2 under the group action, The stabilizer of s in G is i.e., stab(s) {gG|s.(g) = s}. The stabilizer subgroup As we've seen, elements in the same orbit can have different stabilizers. Proposition (HW exercise) Set elements in the same orbit have conjugate stabilizers: stab(s.$(g))=gstab(s)g. for all gЄ G and s € S. In other words, if x stabilizes s, then g¹xg stabilizes s. (g). Here are several ways to visualize what this means and why. Prove the following: Stabc() {gGg-2=2} The stabilizer of z is a subgroup of G. If G acts transitively on X, then the size of the orbit of a is given by G-z]=[G: Stab()], where [G: Staba (z)] denotes the index of the stabilizer subgroup in G. Here is the action graph of the group D₁ = {r, f): 2 Hr³ H+4 labeled by destinations Hr Н labeled by stabilizers 0(x) 5 $(1) (9) (9) $(r) (r) (fr)=(1) (124) In other words, if x is a loop from s, and ss, then g¹xg is a loop from s'. 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).
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter3: Groups
Section3.4: Cyclic Groups
Problem 3E: Exercises
3. Find the order of each element of the group in Example of section.
Example 3.
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