Question #201ff

1 Answer
Feb 24, 2018

Answer:

# "Yes, the text is correct, and (with proper notational set-up)" #
# "is essentially immediate." #

Explanation:

# "What the text said is correct; but this fact can appear" #
# "obscured, I think, due to some notation the text may have left" #
# "out." #

# "So we have:" \ \ G \ \ "is a group," \ \ S = \mathcal{P}(G) \ \ "is the power set of" \ \ G, "and we have an action of" \ \ G \ \ "on" \ \ S \ "-- let's call the action" \ \ \alpha, "defined as follows:" #

# \qquad \qquad \qquad \qquad \qquad \alpha: G xx S rarr S, \qquad \qquad \alpha: ( g, A ) \mapsto g A g^{-1}. \qquad \qquad \qquad (I) #

# "Can check that" \ \alpha \ \ "is an action of" \ \ G \ \ "on" \ \ S. #

# "The normalizer of" \ \ A \ \ "in" \ \ G, "written" \ N_G(A), "is exactly as you" #
# "have defined it:" #

# \qquad \qquad \qquad \qquad \qquad \qquad N_G(A) = \{ g \in G | gA = Ag \}. #

# "Another way it can be defined (I like this one), is" #

# \qquad \qquad \qquad \qquad \quad \quad \ N_G(A) = \{ g \in G | gAg^{-1} = A \}. \qquad \qquad \qquad \qquad \qquad \ (II) #

# "They are equivalent, trivially. I like this one because it targets" #
# "the idea of group-theoretic conjugation:" #

# "normalizer = set of elements of the group that leave the subset" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad "fixed under conjugation." #

# "Makes it a litttle easier to think about, perhaps; but stays with" #
# "the very fundamental idea of conjugation in groups." #

# "But anyway -- I have gotten a little distracted !! Back to your" #
# "question !!" #

# "So we have the action defined in (I), and an (alternative)" #
# "definition of normalizer in (II). Additionally, we have the" #
# "stabilizer of a subset" \ A sube S, "which is the set of elements of" \ \ G, "that leave" \ A \ \ "fixed under the action" \ \ \alpha:" #

# \qquad "stabilizer of" \ \ A \ \ "in" \ \ G \ = \ G_A \ = \ \{ g \in G | \alpha( g, A ) = A \}. #

# "In the case of our action," \ \alpha( g, A ) = gAg^{-1}, \ "this reduces to:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \quad G_A \ = \ \{ g \in G | gAg^{-1} = A \}. #

# "This is exactly the normalizer of" \ \ A, "as defined in (II) above::" #

# \qquad \qquad \qquad \qquad \qquad \qquad N_G(A) = \{ g \in G | gAg^{-1} = A \}. #

# "So, yes, the text is correct, and (with proper notational set-up)" #
# "is essentially immediate." #

# "If I may, I think I see two sources for your confusion:" #

# "1) The definition you gave for the stabilizer is not correct." #
# \qquad \quad "In the notation you used, you gave: " #

# \qquad \qquad \qquad "stabilizer of" \ \ A \ \ "in" \ \ G #

# \qquad \qquad \qquad \quad \quad = \ \{ g \in G | g = gA \ \ "(in the notation you had)" \}. #

# \qquad "In the notation you used, it is: " #

# \qquad \qquad \qquad "stabilizer of" \ \ A \ \ "in" \ \ G #

# \qquad \qquad \qquad = \ \{ g \in G | A = gA \ \ "(in the notation you had)" \}. #

# \qquad "(Note the difference.)" #

# "From the beginning, keep in mind that, in the stabilizer," #
# "what is being stabilized (left fixed) is the subset" \ A sube S. #

# "Also, in the language I think you used, you interpreted the" #
# "stabilizer as:" \qquad "the set of elements of" \ \ G,"where" \ A \ \ "acts" #
# "like an identity on" \ \ G. \ \ "This is not to criticize -- this is a" #
# "good way of being able to think of things. From this point of" #
# "view, the interpretation would be:" \qquad "the stabilizer is the set of" #
# "elements of" \ \ G,"that act like an identity on" \ A. #

# "2) Unless basic hygiene is applied, notation for actions can" #
# "quickly result in confusion, or obfuscation of what's happening." #
# "For example, in the unfortunate standard action notation, the" #
# "action of the element" \ g \in G, "on the subset" \ \ A sube S, "is [crazy]" #
# "denoted:" \ \ gA. \ \ "It looks to all the world like group" #
# "multiplication, but is not meant to be. For example, with this" #
# "notation, the particular action we have been using here would" #
# "be written:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad gA \ = \ gAg^{-1}. #

# "This is crazy. Who would not be confused ?!! The LHS of the" #
# "above uses action operation, and the RHS of the above uses" #
# "group multiplication -- impossible to tell apart." #