Question #201ff

1 Answer
Feb 24, 2018

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

Explanation:

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