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Say $h$ is related to $g$ in some way, denote this by $h:R:g$. Say we also have some kind of operation that we can perform on both $h$ and $g$. Denote this operation by $*$. The relation, $R$, is such that:

$$hRg \implies h*x :R: g*x$$

For example, if $h, g, x \in \mathbb{Z}$, the relation $~\geq~$ with the operation $[\text{multiplication}]$ does not have this property, because one could multiply by $-1$. It would if $h, g, x \in \mathbb{Z}^+$. Equality has this property with any operation (as does any equivalence relation, I think).

Is there a name for this property?


I understand this is somewhat abstract, and I may be ignorant of a whole slew of terminology. To put this in some context, this question came from my having to prove to myself that, if $H$ is a normal subgroup of $G$, then $g*H*g^{-1} = H ~~\forall g \in G$ (where $*$ in this case is the group operation). We certainly know $g*H*g^{-1} \subseteq H$. Group-operating by $g$ on the right and $g^{-1}$ on the left, assuming $(\subseteq, *)$ has this property, would show $H \subseteq g^{-1}*H*g = g*H*g^{-1}$.

(Indeed, $(\subseteq, *)$ has this property)

  • In the motivating example, it's somewhat hard to be certain what $\ast$ refers to, since that notation isn't used anywhere else in that example. Is it "conjugation by $g$ (or $g^{-1}$)" ? So your assertion that "$(\subseteq, \ast)$ has this property" is the claim that $H \subseteq K \implies g^{-1}Hg \subseteq g^{-1}Kg$ ? (As an aside, it may be better to call your operation $f$ or something, so you can write $f(h) : R : f(g)$ above; while $\ast$ is OK, I'm not sure what $x$ has to do with anything) – pjs36 Feb 03 '18 at 19:06
  • @pjs36 Check out my edit (it was made 3-seconds after your comment, haha). Does this make more sense? (You're right about not-using-this-notation anywhere else. I put it into my group example to be more explicit.) – AmagicalFishy Feb 03 '18 at 19:08
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    Yes, that does help, I think. I guess in the example at least, that $h, g$ belong to a set $S$ on which a relation $R$ is defined, and that you have a group acting on $S$ (at first, I thought all of $h, g$, and $x$ would be coming from $S$, but that's not the case). – pjs36 Feb 03 '18 at 19:27

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