Cyclic four-subgroups of symmetric group:S4

This article is about a conjugacy class (also an equivalence class up to automorphisms) in the symmetric group of degree four, comprising cyclic groups of order four.

We denote the symmetric group $$S_4$$ on $$\{ 1,2,3,4 \}$$ by $$G$$. We define:

$$H := \langle (1,2,3,4) \rangle = \langle (1,4,3,2) \rangle = \{, (1,2,3,4), (1,3)(2,4), (1,4,3,2) \}$$.

Its other images under inner automorphisms (which are the same as its images under automorphisms) are:


 * $$\langle (1,3,2,4) \rangle = \langle (1,4,2,3) \rangle = \{, (1,3,2,4), (1,2)(3,4), (1,4,2,3) \}$$.
 * $$\langle (1,3,4,2) \rangle = \langle (1,2,4,3) \rangle = \{, (1,3,4,2), (1,4)(2,3), (1,2,4,3) \}$$.

Isomorph-conjugate subgroup
$$H$$ is an isomorph-conjugate subgroup of $$G$$: it is conjugate to all the other subgroups isomorphic to it. Hence, it is also an isomorph-automorphic subgroup and an automorph-conjugate subgroup. Note that since $$G$$ is a complete group (symmetric groups are complete) all subgroups of $$G$$ are automorph-conjugate.

Subgroup whose join with any distinct conjugate is the whole group
The join of $$H$$ and any conjugate of $$H$$ distinct from $$H$$ is the whole group. In particular, this forces that:


 * $$H$$ is a satisfies property::pronormal subgroup and hence also a satisfies property::weakly pronormal subgroup.
 * Since $$H$$ is an automorph-conjugate subgroup, this also forces that $$H$$ is a satisfies property::procharacteristic subgroup and satisfies property::weakly procharacteristic subgroup.

Opposites satisfied

 * $$H$$ is a satisfies property::core-free subgroup of $$G$$: the normal core of $$H$$ in $$G$$ is trivial.
 * $$H$$ is a satisfies property::contranormal subgroup of $$G$$: the normal closure of $$H$$ in $$G$$ is $$G$$.

Opposites dissatisfied

 * $$H$$ is not a dissatisfies property::self-normalizing subgroup of $$G$$.
 * $$H$$ is not an dissatisfies property::abnormal subgroup of $$G$$ or a dissatisfies property::weakly abnormal subgroup of $$G$$, because both of these would imply that $$H$$ is a self-normalizing subgroup of $$G$$.