Krdytkyu

The dihedral group of order $2n$ with $n$ odd has property that the complement of cyclic subgroup of order $n$ is a single conjugacy class.

Q. Are there any other solvable groups in which complement of a normal subgroup is a single conjugacy class?

There are solvable groups in which every non-trivial coset of a fixed normal subgroup is conjugacy class; but my consideration is for single conjugacy class outside a normal subgroup.

Let $N$ be any finite solvable group equipped with an action of a cyclic group $C={1,a}$ of order $2$ which fixes no elements besides the identity. For instance, $N$ could be any abelian group of odd order (not necessarily cyclic) with $a$ acting by negation. Let $G=Nrtimes C$ be the semidirect product given by this action. Then $G$ is solvable, $N$ is a normal subgroup, and I claim that $Gsetminus N$ is a single conjugacy class. Indeed, note that the element $(1,a)in G$ does not commute with $(x,1)$ for any $xneq 1$ in $N$ (since $a$ acts nontrivially on every non-identity element of $N$). It follows that the centralizer of $(1,a)$ is just ${(1,a),(1,1)}$, and so the conjugacy class of $(1,a)$ contains $|G|/2$ elements. But this conjugacy class is a subset of $Gsetminus N$ which also has $|G|/2$ elements, so the conjugacy class must be all of $Gsetminus N$.

Conversely, all finite examples are of this form: suppose $G$ is a finite group with a normal subgroup $N$ such that $Gsetminus N$ is a conjugacy class. Picking any $ain Gsetminus N$, we then have $|Gsetminus N|=|G|/|C_G(a)|$. In particular, since $|C(x)|geq 2$, we have $|Gsetminus N|leq |G|/2$ and so $|N|geq|G|/2$. This can only happen if $|N|=|G|/2$, so $N$ must have index $2$ and $C_G(a)$ must have order $2$.

This means that $a$ must be an element of $G$ of order $2$ which commutes with no elements of $G$ except itself and $1$. Moreover, we see that $G$ is the semidirect product of the subgroup $C$ generated by $a$ and $N$, and $a$ must act nontrivially on every non-identity element of $N$ since $a$ commutes with nothing except itself and $1$.