Let $G$ be a group and $x \in G$. The conjugacy class $[x]$ is also the adjoint-orbit $\mathcal{O}_x \equiv \{gxg^{-1} | \text{all } g \in G\}$. The centralizer of $x$ is then the stablizer subgroup $G_x$ with respect to the adjoint action.
We know that the elements of $[x] = \mathcal{O}_x = \{x_0 = x, x_1 = g_1 x g_1^{-1}, …, x_{n-1} = g_{n-1}x g_{n-1}^{-1}\}$ are in one-to-one-correspondence with the cosets $G_x, g_1G_x, … g_{n-1}G_x$. However, I'm curious about where the elements of $[x]$ sit with respect to these cosets. Here are some first thoughts.
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$x \in G_x$.
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Obviously, $x_i$ doesn't sit in $g_i G_x$, despite the correspondence. Otherwise, there exists a $g \in G_x$, such that $g_i x g_i^{-1} = g_i g$, which implies $g^{-1}x = g_i$; but this is impossible, since $g_i$ is not supposed to be inside $G_x$. However, $x_i$ must sit in some other coset $g_j G_x$. Can two different $x_i$ sit in one coset?
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Also obviously, $x_i \in g_i G_x g_i^{-1}$. But I wonder: are these mutually-conjugate subgroups $g_iG_x g_i^{-1}$ distinct? Do they "intersect"? If they intersect, are the $x_i$ the intersections? ($g_iG_x$ are certainly distinct.)
Best Answer
Take the Heisenberg group of order $p^3$, $p$ an odd prime, $$G=\langle x,y,z\mid x^p=y^p=z^p=1, yx=xyz, xz=zx, yz=zy\rangle.$$ The center of $G$ is exactly $\langle z\rangle$, which is also the commutator subgroup, with $z=[y,x]= y^{-1}x^{-1}yx$. The group is nilpotent of class two, so the commutator bracket is bilinear.
An element $x^ay^bz^c$ commutes with $x$ if and only if $$1=[x^ay^bz^c,x] = [x,x]^a[y,x]^b[z,x]^c = [y,x]^b = z^b$$ which holds if and only if $p|b$. Thus, the centralizer of $x$ is $\langle x,z\rangle$.
A complete set of coset representatives is therefore given by $e$, $y$, $y^2,\ldots,y^{p-1}$.
So the conjugates of $x$ are precisely the elements of the form $$y^{i}xy^{-i} = x(x^{-1}y^{i}xy^{-i}) = x[x,y^{-i}] = x[x,y]^{-i} = xz^i$$ with $0\leq i\lt p$.
All of these elements are in $C_G(x) = G_x$; so they all lie in the exact same coset of the centralizer. So the answer to question 1 is "yes."
As Brauer Suzuki noted in comments, $g_iG_xg_i^{-1} = G_{g_ixg_i^{-1}}$, so the intersection is the centralizer of the subgroup generated by $[x]$. This always includes the center, so in many cases, this is nontrivial.
In the example above, it is precisely $\langle x,z\rangle = G_x$, because $G_x$ is normal (as it contains the commutator subgroup).