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The definition of normalizer of a non-empty set $A$ in group $G$ is given by $$N_G(A) = \{g\in G\:|\:gAg^{-1} = A\}$$

This definition is from Abstract Algebra by Dummit and Foote.

Another definition I have seen around the internet (here and in this video) is $$N_G(A)=\{g\in G\:|\:gag^{-1}\in A\:\forall\: a\in A\}$$

I try to prove these equivalent. So we need $$gAg^{-1}=A \iff gag^{-1}\in A \:\forall\: a \in A$$

Here is my partial proof:

Assume $gAg^{-1}=A$. Then $gag^{-1}\in gAg^{-1}=A \:\forall \:a \in A$.

Conversely, assume $gag^{-1}\in A \:\forall\: a \in A$. Let $x\in gAg^{-1}$. Then, $\exists\:a_x \in A$ such that $x=ga_xg^{-1}\in A$. It follows that $gAg^{-1}\subseteq A$.

Now , let $y \in A$. This implies $gyg^{-1} \in A$. (This is where I'm stuck).

If I could somehow prove $y \in A \implies g^{-1}yg\in A$, then the result would from the hypothesis. Please let me know if I'm headed in the wrong direction.

I have seen this MathSE post. The definition of the normalizer is a slightly modified version of the first one stated here. However, I would like to know how I can proceed from where I'm stuck.

DS2830
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  • That's the correct way. Simply replace $g$ with $g^{-1}$ in the definition, after you've proved the normaliser is a subgroup. – Bernard Oct 19 '19 at 13:26
  • @Bernard I am not entirely convinced by this. I am thinking that we let $t = g^{-1}$ and use $t$ instead of $g$ in the definition. However, this seems a little flaky to me. Can you provide a proof of why this replacement is allowed? – DS2830 Oct 19 '19 at 13:41
  • Are you supposed to know that the normaliser is a subgroup (in both sense)? – Bernard Oct 19 '19 at 13:46
  • @Bernard I know normalizer is a subgroup but only in the first sense. – DS2830 Oct 19 '19 at 13:47
  • @Bernard I mean the first definition. – DS2830 Oct 19 '19 at 13:48
  • By the way, you might read that MathSE post you linked a little more carefully. The accepted answer is wrong, and the right answer is the unaccepted one. – Lee Mosher Oct 19 '19 at 15:19
  • And now I'm going to delete my own answer, because this question is a duplicate. – Lee Mosher Oct 19 '19 at 15:20
  • @LeeMosher You reduced my problem to $gAg^{-1}=A \iff gAg^{-1} \subset A$. I don't understand why on the right hand side, the subset needs to be proper. I mean isn't it obviously false that $A \subset B \implies A = B$. Shouldn't we have $\subseteq$ instead of $\subset$? – DS2830 Oct 19 '19 at 15:42
  • You can read that (deleted) answer with $\subseteq$ in place of $\subset$, if you like. It is common in set theory for only $\subset$ to be used and for $\subseteq$ to be unused. – Lee Mosher Oct 19 '19 at 15:48
  • I am a fool. This post: https://math.stackexchange.com/questions/3000680/definitions-of-normalisers-for-infinite-groups?rq=1 covers my problem exactly. – DS2830 Oct 19 '19 at 16:12
  • @LeeMosher I appreciate the time and effort you took to solve my problem but I wonder why you deleted it. It was a fantastic answer and believe it could be of help to many if it stayed up. Could you please repost it if it is not too much of a hassle? – DS2830 Oct 19 '19 at 16:36
  • I deleted it because it turns out to be the exact same answer as is given in those earlier problems. – Lee Mosher Oct 19 '19 at 16:51

1 Answers1

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$gAg^{-1}$ = {$gag^{-1} : a\in A$}

Therefore, $gAg^{-1} = A$ occurs iff $g^{-1}Ag=A$ which both imply $\forall a\in A\; gag^{-1},g^{-1}ag\in A$

Now, assume $\forall a\in A\; gag^{-1}\in A$. From this we have $gAg^{-1}\subset A$. Choose $b\in A$ and note that $y=g^{-1}bg\in A$ and therefore $$b=gyg^{-1}\in gAg^{-1}$$

implying $A\subset gAg^{-1}$. Therefore, $gAg^{-1} = A$ iff $\forall a\in A\; gag^{-1}\in A$

Oliver Kayende
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