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Given a group $G$ and a subset of the group $S$, the normalizer $N_G(S)$ is defined as the set $$ N_G(S) = \{ g\in G \mid \forall s\in S, \ gsg^{-1}\in S \} $$

Is a weaker version of that (which is certainly not a group), such that $$ W_G(S) = \{ g\in G \mid \exists s\in S, \ gsg^{-1}\in S \} $$ well-known and studied?

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    $\begingroup$ What do you mean by "is there... ?", do you mean it in a "was that ever studied?" way perhaps? $\endgroup$
    – Bruno B
    Commented May 8, 2023 at 20:57
  • $\begingroup$ Yes. Is this concept well understood, does it have nice properties, interesting applications, etc. Basically I just want an answer of the type, "yes, this is called the ____ , see source ____". $\endgroup$
    – Undead
    Commented May 8, 2023 at 21:23
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    $\begingroup$ If $e\in S$, then $W_G(S)=G$. $\endgroup$
    – Shaun
    Commented May 8, 2023 at 21:54
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    $\begingroup$ Extending the comment of @Shaun : if $S \cap Z(G) \neq \emptyset$, then $W_G(S)=G$. $\endgroup$ Commented May 8, 2023 at 22:18
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    $\begingroup$ To generalize the observation even more, if $S$ contains a conjugacy class, then $W_G(S)=G$. this includes the case of $S$ containing a central element, since central elements are their own conjugacy class. $\endgroup$ Commented May 9, 2023 at 1:03

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