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Let $K = \cap_{i \in I} K_i$ for normal set subgroups $K_i$.

Most of the proofs I found online just proved one side: $gKg^{-1}\subset K$ and left the other side untouched. For example, the one with a lot of upvotes is: Show that the intersection of two normal subgroup of $G$ is normal subgroup of $G$.

Now, if G is finite, it is not hard to prove $ K = gKg^{-1} $ by observing that the map from K to $gKg^{-1}$ is injective. However, if K is infinite, such result may not hold. Examples can be found here: Conjugate subgroup strictly contained in the initial subgroup?

This has left me in deep confusion. If the cardinality of K is not specified, does this result hold?

Shaun
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Dinoman
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  • What is your definition of a normal subgroup? Isn't it that $K$ is a normal subgroup of $G$ if for any $k\in K, g\in G$ we have $gkg^{-1}\in K$? If so there is nothing more you need to show. – Snaw Feb 25 '22 at 06:43
  • @Snaw The definition I learned is that $gKg^{-1}=K$, which at first is a stronger condition. – Vercassivelaunos Feb 25 '22 at 07:41
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    As you will have seen from the answer below, the point lies with the quantifier. If $g K g^{-1} \subseteq K$ for all $g \in G$, then $g K g^{-1} = K$ for all $g \in G$. – Andreas Caranti Feb 25 '22 at 08:15

1 Answers1

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If $gKg^{-1}\subset K$ for all $g$, then for any $g\in G, k\in K$, we have $$h = g^{-1} k (g^{-1})^{-1} \in g^{-1} K g \subset K$$ hence $k = ghg^{-1}\in gKg^{-1}$.

Just a user
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