In Group Theory, what is the difference between a normalizer and a centralizer of a set S? I'm having a bit of difficulty understanding it...

Thanks in advance!

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    $\begingroup$ Have you looked at the Wikipedia page centralizer and normalizer? If so, can you specify what it is that you don't understand from the page? $\endgroup$ – Zev Chonoles Aug 15 '11 at 19:56

There is a difference between fixing $S$ elementwise and preserving $S$ while possibly permuting its elements — you may have come across this distinction before, in linear algebra. To be precise, the centralizer of $S$ in an ambient group $G$ is $$ Z_S = \{x \in G : \text{for each } y \in S,\, xyx^{-1} = y\} $$ and the normalizer is $$ N_S = \{x \in G : xSx^{-1} = S\}, $$ where $xSx^{-1} = \{xyx^{-1} : y \in S\}$. Certainly $Z_S \subset N_S$, but this containment of subgroups may be strict. For a simple example, take $G = S_3$ and $S = A_3 = \{e, [123], [132]\}$. Here $Z_S = S$, but $N_S = G$.

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  • $\begingroup$ Ah-hah! I hadn't even considered the possibility of permuting elements...Thanks so much! :) $\endgroup$ – Dan M. Katz Aug 15 '11 at 20:07

If $G$ is a group, and $H$ is a subgroup, then the normalizer of $H$ in $G$ is $$N_G(H) = \{ g\in G \mid g^{-1}Hg = H\},$$ and the centralizer is $$C_G(H) = \{g\in G \mid gh = hg\text{ for all }h\in H\}.$$

It is easy to see that $C_G(H)\subseteq N_G(H)$, but the converse need not hold.

For example, take $G=S_3$, and let $H = \{ I, (1,2,3), (1,3,2)\}$.

What is $C_G(H)$? It's the collection of all permutations that commute with $I$, with $(1,2,3)$, and with $(1,3,2)$. Since $(1,2)$ does not commute with $(1,2,3)$, $$(1,2,3)(1,2) = (1,3)\neq (2,3) = (1,2)(1,2,3),$$ then $(1,2)\notin C_G(H)$. However, $(1,2)$ does normalize $H$: $$\begin{align*} (1,2)^{-1}I(1,2) &= I\in H;\\ (1,2)^{-1}(1,2,3)(1,2) &= (1,3,2)\in H;\\ (1,2)^{-1}(1,3,2)(1,2) &= (1,2,3)\in H. \end{align*}$$ So $(1,2)\in N_G(H)$. Similarly, $(1,3)$ and $(2,3)$ are not in the centralizer, but are in the normalizer. $H$ is contained in both.

For another example, take $G=H=S_3$. Then the normalizer is all of $G$, because for every $x,g\in G$ we have $gxg^{-1}\in G$; but the centralizer is equal to the center (the set of things that commute with everything) and the center of $G$ is just the identity.

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  • $\begingroup$ Are you sure about C_G(H)\subseteq N_G(H) ? $\endgroup$ – I.R Jan 18 '17 at 10:25
  • $\begingroup$ @Arturo Magidin Normalizer and centralizer of a subset $H$ of a group $G$ are different as you have clearly explained, but it seems like normalizer and centralizer of an element $a$ of $G$ are same. Isn't it? $\endgroup$ – gete Sep 10 '19 at 16:40
  • $\begingroup$ @get: $xhx^{-1}=h$ if and only if $xh=hx$. That's why nobody talks about "normalizers" of an element (or, more generally, seldom of a subset; in general we talk about normalizers for subgroups). $\endgroup$ – Arturo Magidin Sep 10 '19 at 16:51
  • $\begingroup$ @Arturo Magidin Ok. Thanks a lot for clarification $\endgroup$ – gete Sep 10 '19 at 17:39

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