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The definition I am using is $$N_G(K) = \{ g \in G : gkg^{-1} \in K \ \forall k\in K \} $$

I am trying to show that if the element $g\in N_G(K)$ then $g^{-1} \in N_G(K)$.

Proof Idea: If $k_1\in K$ our claim is that $g^{-1}k_1g \in K$. I can seem to get from the fact that $gkg^{-1} \in K$ for all $k\in K$ to $g^{-1}k_1g \in K$ thus I am stuck.

Is what I am trying to prove even true ?

I saw this proof that is similar to mine but it is wrong(I think so):

Let $g$ be in $N_G(K)$.

Note that for any $k$ in $K$, $g^{(-1)} k (g^{(-1)})^{(-1)} = (g k^{(-1)} g)^{(-1)}$. (Wrong identity thus proof fails)

Since $g$ is in $N_G(K)$, $g k^{(-1)} g$ is in $K$ (because $k^{(-1)}$ is in $K$), and thus $(g k^{(-1)} g)^{(-1)}$ is also in $K$.

Therefore, $g^{(-1)}$ is also in $N_G(K)$.

Please use this definition when answering the question and tell me why the proof above fails.

Nameless
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  • Possible duplicate? https://math.stackexchange.com/questions/1407012/proof-that-normalizer-and-center-are-subgroups – Gregory Oct 19 '17 at 00:52
  • @Gregory I read that post before posting this one but didn't find the answer in the notation written above. – Nameless Oct 19 '17 at 00:57
  • They write $N = {g \in G : gS = Sg }$ replace $S$ with $K$ and it looks the same to me. – Gregory Oct 19 '17 at 01:06
  • $N = { g \in G : gK = Kg } = {g \in G : g k g^{-1} \in K, k \in K }$, does it look the same now? – Gregory Oct 19 '17 at 01:07
  • @Gregory doesn't the second set(after 3rd equality sign) have a for all quantifier on $k$ while the first one has a there exists quantizer on $k\in K$ ? – Nameless Oct 19 '17 at 01:16
  • Please don’t put ever the quantifiers after the condition. Doing so can cause ambiguity. – Lubin Oct 19 '17 at 01:54
  • As Lubin's answer shows, your definition is incorrect when the group is infinite. You can find other counterexamples here : https://math.stackexchange.com/questions/107862/conjugate-subgroup-strictly-contained-in-the-initial-subgroup, https://math.stackexchange.com/questions/2413065/alternative-definition-for-normalizer/2413082#2413082 – Arnaud D. Oct 19 '17 at 09:31
  • @ArnaudD. Does it work when the group is infinite ? – Nameless Oct 19 '17 at 13:35
  • @Gregory I checked but the two ${g \in G: gKg^{-1}= K } = { g\in G: gK = Kg} \neq {g\in : gkg^{-1} \in K, k\in K}$. – Nameless Oct 19 '17 at 13:37
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    @Nameless When the subgroup $K$ is finite, since conjugation by $g$ is a bijection, $gKg^{-1}$ and $K$ have the same number of elements. So $gKg^{-1}\subset K$ implies $gKg^{-1}=K$. – Arnaud D. Oct 19 '17 at 13:48

1 Answers1

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The reason you can’t prove it is that the “definition” you’re using is wrong.

What you are using is $N(K)=\{g\in G: \forall k\in K, gkg^{-1}\in K\}$, or in other words, $N(K)=\{g\in G: gKg^{-1}\subset K\}$. But the right definition is $$ N_G(K)=\{g\in G:gKg^{-1}=K\}\,. $$ Here’s a (fairly recondite) example that gives a case where the set defined in your formulation is not closed under inverse: for the underlying sets, take $\Bbb Z\times\Bbb Q$, but with the unusual law of combination depending on a fixed positive integer $p$ (need not be prime) $$ (m,a)\star(n,b)=(m+n,a+p^mb)\,. $$ If I have my group theory right, this is an example of a semidirect product. At any rate, you see that no matter how you parenthesize the three-fold product, you get $$ (m,a)\star(n,b)\star(r,c)=(m+n+r,a+p^mb+p^{m+n}c)\,. $$ So, this law of combination is associative, clearly noncommutative, and the $\star$-inverse of $(m,a)$ will be $(-m,-a/p^m)$. Thus we have a good group. Let’s look at the set $K$ of all $(0,k)$ for $k\in\Bbb Z$. You see that it’s a good subgroup, isomorphic to the additive group $\Bbb Z$. Now, according to your formulation, $g=(1,0)$ is in the pseudo-normalizer of $K$, since for every $k$, $g\star(0,k)\star g^{-1}=(1,0)\star(0,k)\star(-1,0)=(1,0)\star(-1,k)=(0,pk)\in K$. But you check that $g^{-1}\star(0,k)\star g=(0,k/p)\notin K$, certainly not for every $k$, anyway.

Lubin
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