# Equivalent definitions for a Normalizer of a Subgroup

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?

• 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? Oct 11, 2020 at 21:22
• 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). Oct 11, 2020 at 21:53
• Related Oct 11, 2020 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? Oct 11, 2020 at 22:46
• @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. Oct 12, 2020 at 15:32

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: $$H\triangleleft G\iff N(H)=G$$.
• For the $\text{Cute one}$ (+1),,,:)) Oct 12, 2020 at 8:00