If a group centralizer is defined as $C_G(A)=\{g \in G : gag^{-1} = a$ for all $a \in A\}$, and a group normalizer is defined as $N_G(A)=\{g\in G:gAg^{-1}=A\}$, where $gAg^{-1}=\{gag^{-1}:a\in A\}$ (definition taken from Abstract Algebra by Dummit and Foote), then what's the difference between $C_G(A)$ and $N_G(A)$?

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    $\begingroup$ I think maybe I'm getting it. There could be some elements in $G$ such that $gag^{-1} \neq a$, yet $gag^{-1}$ defines some element in $A$. $\endgroup$
    – sequence
    Sep 27, 2015 at 21:44
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    $\begingroup$ Does this answer your question? Normalizer vs Centralizer $\endgroup$ Sep 11, 2021 at 17:06

3 Answers 3


I would say (less precisely, but correctly) like this:

  • $g$ is in $N_G(A)$ means $gag^{-1}=$ some $a'$ in A ($a\in A$).

  • $g$ is in $C_G(A)$ means $gag^{-1}=$ same $a$ in A ($a\in A$).

We should note that although there is difference between these two notions, there is also a relation between them: $$C_G(A) \mbox{ is always contained in } N_G(A).$$

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    $\begingroup$ We can get even better, $C_G(A)$ is a normal subgroup of $N_G(A)$. $\endgroup$ Jul 8, 2019 at 14:28
  • $\begingroup$ very well put!! I always waned to know the difference and now here it is clear cut and concise! $\endgroup$
    – anonymous
    Oct 15, 2022 at 17:07

Consider $G$ acting by conjugation on itself. Elements of $N_G(A)$ restrict to permutations of $A$, and more particularly elements of $C_G(A)$ restrict to the identity permutation of $A$.


Let H is a Subgroup of G. Now if H is not normal if any element ${g \in G}$ doesn't commute with H. Now we want to find if not all ${g \in G}$, then which are the elements of G that commute with every element of H? they are normalizer of H. i.e., the elements of G that vote 'yes' for H when asked to commute.

  • Hence, ${N_G(H)=\{g \in G: gH=Hg }\}$
  • | Now Centralizer of an element ${a \in G}$ is set of all elements that commutes with ${a}$. in layman's term element ${a}$ looks for the element in G that commutes with it.
  • Hence ${C(a)=\{g\in G : ag=ga}\}$

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