# Question #201ff

Feb 24, 2018

$\text{Yes, the text is correct, and (with proper notational set-up)}$
$\text{is essentially immediate.}$

#### Explanation:

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

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

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \alpha : G \times S \rightarrow S , \setminus q \quad \setminus q \quad \setminus \alpha : \left(g , A\right) \setminus \mapsto g A {g}^{- 1} . \setminus q \quad \setminus q \quad \setminus q \quad \left(I\right)$

$\text{Can check that" \ \alpha \ \ "is an action of" \ \ G \ \ "on} \setminus \setminus S .$

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

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {N}_{G} \left(A\right) = \setminus \left\{g \setminus \in G | g A = A g \setminus\right\} .$

$\text{Another way it can be defined (I like this one), is}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus {N}_{G} \left(A\right) = \setminus \left\{g \setminus \in G | g A {g}^{- 1} = A \setminus\right\} . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \left(I I\right)$

$\text{They are equivalent, trivially. I like this one because it targets}$
$\text{the idea of group-theoretic conjugation:}$

$\text{normalizer = set of elements of the group that leave the subset}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{fixed under conjugation.}$

$\text{Makes it a litttle easier to think about, perhaps; but stays with}$
$\text{the very fundamental idea of conjugation in groups.}$

$\text{But anyway -- I have gotten a little distracted !! Back to your}$
$\text{question !!}$

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

$\setminus q \quad \text{stabilizer of" \ \ A \ \ "in} \setminus \setminus G \setminus = \setminus {G}_{A} \setminus = \setminus \setminus \left\{g \setminus \in G | \setminus \alpha \left(g , A\right) = A \setminus\right\} .$

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

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {G}_{A} \setminus = \setminus \setminus \left\{g \setminus \in G | g A {g}^{- 1} = A \setminus\right\} .$

$\text{This is exactly the normalizer of" \ \ A, "as defined in (II) above::}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {N}_{G} \left(A\right) = \setminus \left\{g \setminus \in G | g A {g}^{- 1} = A \setminus\right\} .$

$\text{So, yes, the text is correct, and (with proper notational set-up)}$
$\text{is essentially immediate.}$

$\text{If I may, I think I see two sources for your confusion:}$

$\text{1) The definition you gave for the stabilizer is not correct.}$
$\setminus q \quad \setminus \quad \text{In the notation you used, you gave: }$

$\setminus q \quad \setminus q \quad \setminus q \quad \text{stabilizer of" \ \ A \ \ "in} \setminus \setminus G$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad = \setminus \setminus \left\{g \setminus \in G | g = g A \setminus \setminus \text{(in the notation you had)} \setminus\right\} .$

$\setminus q \quad \text{In the notation you used, it is: }$

$\setminus q \quad \setminus q \quad \setminus q \quad \text{stabilizer of" \ \ A \ \ "in} \setminus \setminus G$

$\setminus q \quad \setminus q \quad \setminus q \quad = \setminus \setminus \left\{g \setminus \in G | A = g A \setminus \setminus \text{(in the notation you had)} \setminus\right\} .$

$\setminus q \quad \text{(Note the difference.)}$

$\text{From the beginning, keep in mind that, in the stabilizer,}$
$\text{what is being stabilized (left fixed) is the subset} \setminus A \subseteq S .$

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

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

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad g A \setminus = \setminus g A {g}^{- 1} .$

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