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To my dear friends with gratitude.
I want to get help proving centralizer of a nonempty subset of a group is a normal subgroup in the normalizer of that set in the mentioned group.symbolically:
$C_G (S)\lhd N_G (S)$

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    $\begingroup$ What have you tried? There are two "usual" ways to do this. One is simply by writing up the definitions and checking by doing some rewriting of things. The other is by defining a map from $N_G(S)$ to a suitable group such that $C_G(S)$ is the kernel. $\endgroup$ – Tobias Kildetoft Oct 31 '13 at 14:14
  • $\begingroup$ I know the second way you have mentioned,but i want to prove by simple normality definition $\endgroup$ – Aref Oct 31 '13 at 14:48
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Let's be $z \in C_G(S)$ and let's be $n \in N_G(S)$ we want to show that $nzn^{-1} \in C_G(S)$.

If $s \in S$, then $(nzn^{-1})s(nzn^{-1})^{-1}=nz(n^{-1}sn)z^{-1}n^{-1}$. Now $n^{-1}sn \in S$ and so $nz(n^{-1}sn)z^{-1}n^{-1}=n(n^{-1}sn)n^{-1}=s$.

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Of course you can try to prove it directly, by using the definitions of normalizer and centralizer nonetheless there's a different approach to the problem.

Here's a little hint (to prove with the different approach).

Generally a good technique in proving that some subgroup is normal is to show that it's the kernel of some homomorphism, proving that centralizer of a subgroup is normal in the normalizer of the same subgroup can be done in this way.

The trick is to find the right homomorphism, but here's a second hint: every action of group $G$ on a set $X$ is the same as an homomorphism of type $G \to \text{Aut}(X)$, i.e. a permutation representation of the group. The normalizer of a subgroup have a natural action on the same subgroup.

Hope this helps.

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  • $\begingroup$ Hello, to extend on this I was trying to figure out the right homomorphism. I defined mine as $\phi : N_G(H) : \rightarrow Aut(H)$, $g \mapsto \phi (g)(h) := ghg^{-1}$ and the kernel is the centraliser of $H$, but I'm not sure if my homomorphism is well defined. What do you think? $\endgroup$ – NotAbelianGroup Jan 14 '19 at 17:22
  • $\begingroup$ @NotAbelianGroup that's actually a good guess, that's an homomorphism indeed. Why do you think that it is not an homomorphism? $\endgroup$ – Giorgio Mossa Jan 14 '19 at 17:47
  • $\begingroup$ I know that $\phi : G \rightarrow Aut(H)$ is an homomorphism but I wasn't sure that the restriction to $N_G(H)$ was also an homomorphism. I guess it is since $N_G(H)$ is a subgroup of $G$. $\endgroup$ – NotAbelianGroup Jan 14 '19 at 18:01
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    $\begingroup$ The restriction of an homomorphism to a subgroup is always an homomorphism as well! $\endgroup$ – Giorgio Mossa Jan 14 '19 at 18:37

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