I am trying to recall what happened to the conjugacy class and centraliser when we quotient out by a normal subgroup.

In particular, we know from Orbit-Stabiliser that

$|G|=|ccl_G(g)||C_G(g)|$ for every $g\in G$

If we quotient out by a normal subgroup we have that


My question is what was the relationship between $|ccl_G(g)|$ and $|ccl_{G/N}(gN)|$ (or $|C_G(g)|$ and $|C_{G/N}(gN)|$)

Do we need $N$ to be a particular subgroup (like a central subgroup, commutator subgroup, etc) to have a special relationship?

Some thoughts:

Certainly, if $g$ and $hgh^{-1}$ are conjugate elements in $G$, then $(hN)(gN)(hN)^{-1}=hgh^{-1}N$ so $gN$ and $hgh^{-1}N$. The question is if $hgh^{-1}\neq g$ then $hgh^{-1}N=gN$ so $hgh^{-1}g^{-1}\in N$ (So, if $N$ is a derived subgroup then, the conjugacy class all become $1$, i.e. the quotient is abelian. However, what happens in the more general context like if $N$ is a central subgroup like $Z(G)$?)

Second observation if $ccl_G(g)$ is a conjugacy class of size $1$, we can't go any smaller so we know that $|C_{G/N}(gN)|=|C_G(g)|/|N|$

  • $\begingroup$ What is your $ccl_{G}(g)$? $\endgroup$ – user549397 Apr 23 at 7:02
  • $\begingroup$ @user549397 I think the OP means the conjugacy class of $g$ $\endgroup$ – leibnewtz Apr 23 at 7:12
  • $\begingroup$ @leibnewtz Really? Then if $C_G(g)$ is the centralizer of $g$ in $G$, why do we have the first equality? $\endgroup$ – user549397 Apr 23 at 7:19
  • $\begingroup$ Yes, I mean the conjugacy class. Is there something wrong with the first equality? $\endgroup$ – daruma Apr 23 at 9:58
  • $\begingroup$ There are (finitely generated) infinite groups with only two conjugacy classes (that is, where every two non-trivial elements are conjugate). See Corollary 1.2 of Denis Osin, Small cancellations over relatively hyperbolic groups and embedding theorems, Ann. Math. (2010). $\endgroup$ – user1729 Apr 23 at 10:45

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