I am trying to fully understand the connection between a centralizer to conjugacy class.

Starting with the definitions:

Center: $Z(G)=\{z\in G:\forall g\in G, zg=gz\}$ which are all the elements in the group that commute

Centralizer: $C_G(S)=\{g\in G:gs=sg \text{ for all } s\in S\}$ which are all the elements of $g$ that commute with all of the elements subset of $S$

Conjugacy class: $Conj(x)=\{h\in G:\exists g\in G: gxg^{-1}=h \}=\{gxg^{-1}: g \in G \}$ can not explain which elements are in the conjugacy class

Center VS Centralizer: the center takes a group and returns all the elements in the group that commute whereas the Centralizer take a subset of the group and return only that elemnts of the subset which commute

So If for example all the subset commute , let say that the group is cyclic and there for every subset commute, we will have $Z(G)=C_G(G)$

Did I get that right? what are Conjugacy classes and how they differ from the centralizer and the center? is there a simple example for these definitions?

  • 1
    $\begingroup$ Well, perhaps the most important relation, and one you shall need to prove the important Class Equation, is that $$\;\text{Conj}\,(x)=\{x\}\iff\left|\text{Conj}\,(x)\right|=1\iff x\in Z(G)\;$$ .Of course, you can also check that $\;Z(G)\subset C_G(S)\;$ , for every $\;\emptyset\neq S\subset G\;$, but this is rather pretty obvious. What you write in the "Center VS Centralizer" part isn't clear to me, in particular that "takes a group..." part... $\endgroup$
    – DonAntonio
    Feb 21, 2018 at 9:45
  • $\begingroup$ $Z(G) = C_G(G)$ is true in all groups and follows immediately from the definition. $\endgroup$
    – Derek Holt
    Feb 21, 2018 at 9:48
  • $\begingroup$ @DonAntonio the "takes a group" was an attempt to understand it as function, what is the domain, range and co-domain $\endgroup$
    – newhere
    Feb 21, 2018 at 9:48

2 Answers 2


Saying that the center of $G$ are the elements of $G$ that commute makes no sense. Commute with what? The center of $G$ is the set of those elements of $G$ that commute with all other elements of $G$.

Your definition of centralizer is correct. Note that the centralizer of $G$ is precisely the center of $G$ (for every group). And that a group is commutative if and only if its center is the whole group.

As for the conjugacy classes: the conjugacy class of $x$ are all those elements of $G$ of the form $hxh^{-1}$. For instance, if $G$ is commutative, then the conjugacy class of each $x\in G$ is $x$ itself (because then $hxh^{-1}=hh^{-1}x=ex=x$). But if $G$ is the group $S_3$ (the group of the permutations of $\{1,2,3\}$), then you can check that the conjugacy class of $(1\ \ 2\ \ 3)$ is $\{(1\ \ 2\ \ 3),(1\ \ 3\ \ 2)\}$. Note that $x$ always belongs to the conjugacy class of $x$ (because $x=exe^{-1}$).

  • $\begingroup$ He mistakenly said that the centralizer of a subset returns the elements of the subset that commute with all other elements, but he should have said that, given a subset H, the centralizer of H in G returns all elements of G that commute with every element of H. $\endgroup$ Apr 6, 2022 at 14:41

By definition You have $Z(G)=C_G(G)$. If an element is in the center, it commutes with all elements in $G$ as You correctly said and that means for $x\in Z(G)$ that $gxg^{-1}=x$ for all $g\in G$. The mappings $\phi_g:G\rightarrow G,x\mapsto gxg^{-1}$ are called the inner automorphisms of the group $G$ and make the group operate on itself. The conjugacy classes are the orbits of elements under this operation and the elements in the center are exactly those that have orbits of length $1$.

  • $\begingroup$ Can someone please explain the downvote? $\endgroup$ Feb 24, 2018 at 8:53
  • $\begingroup$ ...especially since my answer is not very different from @José Carlos Santos answer (1 minute earlier though) $\endgroup$ Feb 24, 2018 at 11:07

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