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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, 5.0(1) = s. (i) Inverses. That is, if s.4(g)-s, then s.(g¹)=s. (iii) Closure. That is, if s.4(g) =s and 5.0(n)=s, then s.d(gh) = s. Alternatively, it suffices to show that if s.(g) =s and s.d(h) =s, then s.(gh¯¹) = 5. The stabilizer subgroup Here is another example of an action, this time of G = De- Lets be the highlighted hexagon, and H-stab(s). Hr2 Hr (121) (14) The stabilizer of s in G is stab(s) = {g€ G|s.(g) = s}. The stabilizer subgroup Let V be a finite-dimensional vector space over a field F, and let G be a group acting linearly on V. For a vector v € V, define the stabilizer of as Stabe()-(G|-- Prove that Stab(v) is a subgroup of G and that if G is a finite group, then Stab(v) is a finite As we've seen, elements in the same orbit can have different stabilizers group as well. 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.d(g). Here are several ways to visualize what this means and why. Here is the action graph of the group ₁ = (r.): Hr³ H 0(x) (9) (f) (9-1xg) (a) 0(5) $(r) d(r) 0(1¹fr)=0(14) In the beginning of this course, we picked a configuration to be the "solved state," and the gave us a bijection between configurations and actions (group elements).

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, 5.0(1) = s. (i) Inverses. That is, if s.4(g)-s, then s.(g¹)=s. (iii) Closure. That is, if s.4(g) =s and 5.0(n)=s, then s.d(gh) = s. Alternatively, it suffices to show that if s.(g) =s and s.d(h) =s, then s.(gh¯¹) = 5. The stabilizer subgroup Here is another example of an action, this time of G = De- Lets be the highlighted hexagon, and H-stab(s). Hr2 Hr (121) (14) The stabilizer of s in G is stab(s) = {g€ G|s.(g) = s}. The stabilizer subgroup Let V be a finite-dimensional vector space over a field F, and let G be a group acting linearly on V. For a vector v € V, define the stabilizer of as Stabe()-(G|-- Prove that Stab(v) is a subgroup of G and that if G is a finite group, then Stab(v) is a finite As we've seen, elements in the same orbit can have different stabilizers group as well. 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.d(g). Here are several ways to visualize what this means and why. Here is the action graph of the group ₁ = (r.): Hr³ H 0(x) (9) (f) (9-1xg) (a) 0(5) $(r) d(r) 0(1¹fr)=0(14) In the beginning of this course, we picked a configuration to be the "solved state," and the gave us a bijection between configurations and actions (group elements).

Elements Of Modern Algebra
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. We...
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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, 5.0(1) = s. (i) Inverses. That is, if s.4(g)-s, then s.(g¹)=s. (iii) Closure. That is, if s.4(g) =s and 5.0(n)=s, then s.d(gh) = s. Alternatively, it suffices to show that if s.(g) =s and s.d(h) =s, then s.(gh¯¹) = 5. The stabilizer subgroup Here is another example of an action, this time of G = De- Lets be the highlighted hexagon, and H-stab(s). Hr2 Hr (121) (14) The stabilizer of s in G is stab(s) = {g€ G|s.(g) = s}. The stabilizer subgroup Let V be a finite-dimensional vector space over a field F, and let G be a group acting linearly on V. For a vector v € V, define the stabilizer of as Stabe()-(G|-- Prove that Stab(v) is a subgroup of G and that if G is a finite group, then Stab(v) is a finite As we've seen, elements in the same orbit can have different stabilizers group as well. 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.d(g). Here are several ways to visualize what this means and why. Here is the action graph of the group ₁ = (r.): Hr³ H 0(x) (9) (f) (9-1xg) (a) 0(5) $(r) d(r) 0(1¹fr)=0(14) In the beginning of this course, we picked a configuration to be the "solved state," and the gave us a bijection between configurations and actions (group elements).
Transcribed Image Text:Proposition For any 5 € S, the set stab(s) is a subgroup of G. Proof (outline) To show stab(s) is a group, we need to show three things: (i) Identity. That is, 5.0(1) = s. (i) Inverses. That is, if s.4(g)-s, then s.(g¹)=s. (iii) Closure. That is, if s.4(g) =s and 5.0(n)=s, then s.d(gh) = s. Alternatively, it suffices to show that if s.(g) =s and s.d(h) =s, then s.(gh¯¹) = 5. The stabilizer subgroup Here is another example of an action, this time of G = De- Lets be the highlighted hexagon, and H-stab(s). Hr2 Hr (121) (14) The stabilizer of s in G is stab(s) = {g€ G|s.(g) = s}. The stabilizer subgroup Let V be a finite-dimensional vector space over a field F, and let G be a group acting linearly on V. For a vector v € V, define the stabilizer of as Stabe()-(G|-- Prove that Stab(v) is a subgroup of G and that if G is a finite group, then Stab(v) is a finite As we've seen, elements in the same orbit can have different stabilizers group as well. Proposition (HW exercise) Set elements in the same orbit have conjugate stabilizers: stab(s.$(g))=gstab(s)g. for all gЄ G and s€ S. In other words, if x stabilizes s, then g¹xg stabilizes s.d(g). Here are several ways to visualize what this means and why. Here is the action graph of the group ₁ = (r.): Hr³ H 0(x) (9) (f) (9-1xg) (a) 0(5) $(r) d(r) 0(1¹fr)=0(14) In the beginning of this course, we picked a configuration to be the "solved state," and the gave us a bijection between configurations and actions (group elements).
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