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I've seen some definitions of a Normalizer that do not seem equivalent to me and was wondering if I am missing something.

First to setup the basic stuff:

$H$ is a subgroup of $G$.

Denote the normalizer of $H$ as $N(H)$.

So now the different definitions:

Group 1:

  1. $N(H)=\{a \in G: aha^{-1} \in H \text{ for }\forall h \in H\}$
  2. $N(H)=\{a \in G: aHa^{-1} \subseteq H\}$

Group 2:

  1. $N(H)=\{a \in G: aH=Ha\}$
  2. $N(H)=\{a \in G: aHa^{-1}=H\}$
  3. $N(H)=\cup\{Ha : aH=Ha\}$
  4. $N(H)=\cup\{Ha : aHa^{-1}=H\}$

My conclusions so far:

  1. The solutions in Group 1 are equivalent
  2. The solutions in Group 2 are equivalent
  3. The solutions in Group 1 are not equivalent to those in Group 2 in general.
  4. Group 1 basically says to gather all elements for which the conjugate of H is a subgroup of H, while Group 2 says to gather all the elements for which the conjugate of H is equvalent to H.
  5. If G is finite, Group 1 will be equivalent to Group 2, because $aHa^{-1}$ is a bijective image of $H$ so if $H$ is finite and $aHa^{-1} \subseteq H$ => $aHa^{-1}=H$.
  6. So Group 1 and Group 2 are equivalent if G is finite.

So my question comes to this:

Is this correct or not?

Arturo Magidin
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Everstudent
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  • To add some more to this question: It seems to me that the definitions in Group 1 do not make $N(H)$ a group, so unless G is finite in which case Group 1=Group 2 => $N(H)$ is a group. So Group 1 seems like a bad way to define things. So why use the definitions in Group 1 at all? – Everstudent Oct 11 '20 at 21:22
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    The definitions in Group 1 are simply wrong because, as you said, they do not define a subgroup in general. So it is better not to use them at all (unless you are sure that $H$ is finite). – Derek Holt Oct 11 '20 at 21:53
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    Related – Arturo Magidin Oct 11 '20 at 21:58
  • @DerekHolt So why does Pinter use this definition then(Group 1)? Of course he is stating that G is a finite group, so it works, but I still do not see the merit? Isn't it better to just use one of the definitions in Group 2, like Dummit and Foote/Lang/Artin do, so that it works for an infinite group? In general, is there any reason to shy away from normalizers of infinite groups? – Everstudent Oct 11 '20 at 22:46
  • @Everstudent since Pinter is specifically making a definition for finite $G$, you answered your own question about why Pinter used a definition that doesn't work in general. Maybe Pinter learned by experience in teaching finite groups that students are more comfortable with the 1st description of normalizers in a finite group even though it is not a good definition in cases beyond the scope of the book. Should a group isomorphism be defined as a bijective group homomorphism or as a bijective group homomorphism whose inverse is a group homomorphism? For first-time algebra students (contd) – KCd Oct 12 '20 at 00:42
  • I think the first definition is fine, but from the viewpoint of category theory the first definition is "defective" because it is the wrong notion in some non-algebraic categories (e.g., topological spaces). – KCd Oct 12 '20 at 00:43
  • In the question I linked to, in the comments, there is an exchange about the exact phrasing in Pinter. It appears that in an earlier edition it looked at ${x\in G\mid xax^{-1}\in H\text{ if }a\in H}$, but in a later edition this was corrected to ${x\in G\mid xax^{-1}\in H\text{ iff }a\in H}$. Make sure your Definition 1 in Group 1 is correct; I would not that “for $\forall h$” would literally read “for for all $h$”, which is ungrammatical. – Arturo Magidin Oct 12 '20 at 01:46
  • @ArturoMagidin Which later edition are you referring to? I just checked the 2010 edition(I think it's reprinted or something) and it says: ${a \in G: axa^{-1} \in H\ for\ every\ x \in H}$. It does not say ${a \in G: axa^{-1} \in H\ iff\ x \in H}$ – Everstudent Oct 12 '20 at 11:41
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    @Everstudent: I don't have any edition of Pinter. The comments in the question I linked to had that exchange. Take it up with them. Aside: use \text{ for every } to get regular text in the middle of a math formula; otherwise, it comes out awful. – Arturo Magidin Oct 12 '20 at 15:32

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You left one cute one out: the normalizer of $H$ in $G$ is the largest subgroup of $G$ in which $H$ is normal.

Of course, then we have: $H\triangleleft G\iff N(H)=G$.