Conjugacy classes when changing group Let $A \leqslant B \leqslant G$ subgroups, with $[B:A]<\infty$. I would like to understand conjugacy classes when the subgroup with respect to which we consider them changes. Denote $C(a)\in CL(A)$ the conjugacy classes in $A$ and $C(b)\in CL(B)$ the conjugacy classes in $B$ (I hope no confusion will arise). I would like to understand what $A$-classes give rise to a single $B$-class $C(b)$, i.e.
$$\{C(a) \in CL(a) \ : \ \exists x \in B, x^{-1}ax = b\}$$
This implies $x b x^{-1} = a \in A$. This last condition is obviously invariant by left-translation of $x$ by $A$, so that we can afford to be looking at $x \in A/B$. Is the above set in bijection with
$$\{x \in B/A \ : \ x^{-1}bx \in A\}?$$
I can prove it painfully (looking at the natural map $C(a) \mapsto x$ and checking it is injective and surjective. Is there a more elegant way to prove it?
 A: Not exactly clear what you are looking for but perhaps this helps. 
We have $A
\leq B$. Fix an element $b \in B$, and put $T_b=\{ x \in B: xbx^{-1} \in A\}$. This is a subset of $B$ and if it is non-empty we can define an equivalence relation on this set by $x \sim  y$ if and only if $x=yc$ for some $c \in C_B(b)$. (Here $C_B(b)$ denotes the centralizer of $b$, that is, all elements of $B$ that commute with $b$. This is a subgroup by the way). Note that $yc \in T_b$ and indeed $\sim$ is an equivalence relation.
This implies that we can write $T_b$ as a disjoint union as follows: $\bigcup_{i=1}^kx_iC_B(b)$ for some positive integer $k$ and the $x_i \in T_b$ are representatives of the different $k$ equivalence classes. Denote $Cl_B(b)$ as the conjugacy class of $b$ in $B$. Now we can define a map $f: \{x_1, \cdots, x_k\} \rightarrow Cl_B(b) \cap A$ by $f(x_i)=x_ibx_i^{-1}$. This map is well-defined ($x_i \in T_b$ for all $i$) and it is a bijection. Injectivity: if $x_ibx_i^{-1}=x_jbx_j^{-1}$, then $x_j^{-1}x_i \in C_B(b)$, whence $x_i \sim x_j$, so $i=j$. Surjectivity: an element of $C_B(b) \cap A$ is of the form $xbx^{-1} \in A$ for some $x \in B$, so $x \in T_b$, whence $x=x_ic$ for some $i$ and $c \in C_B(b)$. So $f(x_i)=x_ibx_i^{-1}=x_icbc^{-1}x_i^{-1}=xbx^{-1}$. It follows in particular that $Cl_B(b) \cap A=\emptyset$ or $\#(Cl_B(b) \cap A)=k$.
