Conjugacy classes of an element that are the same

Question

Recently I have been studying the transfer homomorphism, and it came to mind that whether conjugacy class of an element with respect to some subgroup is the same as the original group. Namely, if $x \in H - Z(G)$, (where H is a proper nontrivial subgroup of a non abelian finite group $G$ and $Z(G)$ is the center of $G$) then can it be possible that $\mathcal{C}_{H}(x) = \mathcal{C}_{G}(x)$? where $\mathcal{C}_{H}(x)$ and $\mathcal{C}_{G}(x)$ denote the conjugacy classes of x with respect to $H$ and $G$ respectively. Can anyone give me an example of a group $G$ and a subgroup $H$ such that the above mentioned thing can happen?

Answer 1

You can use the symetric groups : take $G= \mathcal{S}_4$, $H = \mathcal{A}_4$ and $x=(12)(34)$. Then,as $\sigma \circ x \circ \sigma^{-1} = (\sigma(1)\;\sigma(2))(\sigma(3)\;\sigma(4))$, we have $$C_G(x)= \{(12)(34),(13)(24),(14)(23)\}.$$

We know that $C_H(x) \subset C_G(x)$. With $\sigma= (2 \; 3 \; 4) \in H$, we have $\sigma \circ x \circ \sigma^{-1} = (1\; 3)(4\;2)=(1\; 3)(2\;4)$ ; and with $\sigma= (2 \; 4 \; 3) \in H$, we have $\sigma \circ x \circ \sigma^{-1} = (1\; 4)(2\;3)$, thus we also have $$C_H(x)= \{(12)(34),(13)(24),(14)(23)\}.$$

Answer 2

Perhaps this characterization gives some clue to finding other examples. Since $cl_H(x)\subseteq cl_G(x)$, then $cl_G(x)=cl_H(x)\iff$ $cl_G(x)\subseteq cl_H(x)$, namely iff, for every $g\in G$, there is a $h\in H$ such that $g^{-1}xg=$ $h^{-1}xh$, namely iff $hg^{-1}x=xhg^{-1}$, namely iff $hg^{-1}\in C_G(x)$, namely iff $h\in C_G(x)g$, namely iff $C_G(x)g\cap H\ne\emptyset$: so, a necessary and sufficient condition for the equality to hold, is that $H$ has nonempty intersection with every right coset of the centralizer of $x$ in $G$.