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Let $A(G)$ be the intersection of all subgroups of finite index of an infinite group $G,*$. I have to show that $A(G)$ is a normal subgroup: $A(G) \lhd G$. Unfinished proof:

Because $A(G)$ is an intersection of subgroups, $A(G)$ itself will be a subgroup of $G$. That means that I can use the following equivalence: \begin{align*} A(G) \lhd G \iff \forall g \in G , \forall h \in A(G):g \ * h \ *g^{-1} \in A(G). \end{align*} So let $g \in G$ and $h \in A(G)$. To prove that $g \ * h \ *g^{-1} \in A(G)$, I have to show that $g \ * h \ *g^{-1} \in D$ for every subgroup $D$ with finite index. So let $D$ be a subgroup of $G$ with finite index (this is possible because $G$ itself is a subgroup with finite index). However, now I am stuck. I cannot seem to prove that $g \ * h \ *g^{-1} \in D$. I know that $h \in D$ because $h \in A(G)$, but appart from that I have used everything I know (I think). I would like to know the next step in my proof (except if there is an error in my reasoning, but I don't know anything about actions).

Einsteinwasmyfather
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  • Think about the action of $G$ by conjugation on subgroups. – Qiaochu Yuan Jan 22 '21 at 11:07
  • Sorry, I don't know what an action is (we skipped that part of the theory in class). So I don't think we need that to solve this problem. – Einsteinwasmyfather Jan 22 '21 at 11:08
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    It's time then to look up what an action is. It will be needed everywhere then anyway. (Sorry, I don't want to be petty, but it sounds like a student of number theory writes"Sorry, I don't know what a prime number is -we skipped that part of the theory in class"). – Dietrich Burde Jan 22 '21 at 11:23
  • Then I would assume that the students in that class would not receive any questions about prime numbers ;) – Einsteinwasmyfather Jan 22 '21 at 11:32
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    But if so, and this seems almost unavoidable, wouldn't it be reasonable just to look it up and learn something more? Like for learning a foreign language, why not look up a new word and learn how to use it? – Dietrich Burde Jan 22 '21 at 11:48

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Every subgroup of finite index contains a normal subgroup of finite index: For $G$ group and $H$ subgroup of finite index, prove that $N \subset H$ normal subgroup of $G$ of finite index exists

And so the intersection of all subgroups of finite index is the same as the intersection of all normal subgroups of finite index.

And the intersection of arbitrary family of normal subgroups is normal, as you can read here: https://proofwiki.org/wiki/Intersection_of_Normal_Subgroups_is_Normal

freakish
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Hint: If $D$ is any subgroup with finite index then $g^{-1}Dg$ is also a subgroup with finite index, and so $h$ belongs to it.

Mark
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  • Genius ! I forgot to mention that I already knew that for every subgroup D with finite index, $gDg^{-1}$ is also a subgroup with finite index, my bad ! Also, I assume it is the same with $g^{-1}Dg$ ? – Einsteinwasmyfather Jan 22 '21 at 11:31
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    Yes, $g^{-1}Dg$ is also conjugation of $D$, just by the element $g^{-1}$ instead of $g$. We have $g^{-1}Dg=g^{-1}D(g^{-1})^{-1}$. – Mark Jan 22 '21 at 11:49
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    By the way, action by conjugation (to learn a new word). – Dietrich Burde Jan 22 '21 at 11:50