Cyclic maximal subgroup of dihedral group:D8

This article discusses the dihedral group of order eight and the cyclic maximal subgroup (the subgroup of order four) of this group. We call the dihedral group $$G$$, and use the following presentation:

$$\! G := \langle a,x \mid a^4 = x^2 = e, xax = a^{-1} \rangle$$

and we set:

$$\! H = \langle a \rangle$$.

In the typical description of $$G$$ as a permutation group:

$$\! G := \langle (1,2,3,4), (1,3) \rangle$$

we have:

$$\! H = \langle (1,2,3,4) \rangle$$



Cosets
The subgroup has index two and is hence a normal subgroup. Its left cosets coincide with its right cosets, and there are exactly two of them: the subgroup itself and the rest of the group:

$$H = \{ e, a, a^2, a^3 \}, G \setminus H = \{ x, ax, a^x, a^3x \}$$

Complements
$$H$$ as four permutable complements, which are also precisely its lattice complements, and none of them is a normal complement.

Intermediate subgroups
There are no properly intermediate subgroups because the subgroup is a maximal subgroup.

As the group of symmetries of a square
If we think of the dihedral group of order eight as the group of symmetries of the square, the cyclic maximal subgroup is the subgroup comprising the rotations, or the orientation-preserving symmetries.

Related interpretations:


 * The dihedral group has a two-dimensional faithful irreducible representation over any field of characteristic not equal to two (see linear representation theory of dihedral group:D8). This representation can be realized as taking values in the orthogonal group. This representation is essentially a generalization of the action on the square. Under any such representation, the cyclic maximal subgroup comprises the elements of determinant one, which are the rotations.
 * When the field has size $$3$$ or $$5$$, this representation surjects onto the orthogonal group. In other words, $$O(2,3)$$ and $$O(2,5)$$ are both isomorphic to the dihedral group. In this case, the cyclic maximal subgroup is $$SO(2,3)$$ or $$SO(2,5)$$.

In the context of different embeddings of the dihedral group
Since the cyclic maximal subgroup is an satisfies metaproperty::isomorph-normal characteristic subgroup, it is in particular an satisfies metaproperty::isomorph-normal coprime automorphism-invariant subgroup, and thus a satisfies metaproperty::fusion system-relatively weakly closed subgroup and thus a satisfies metaproperty::Sylow-relatively weakly closed subgroup. In other words, for any group containing the dihedral group of order eight as a $$2$$-Sylow subgroup, this subgroup is a weakly closed subgroup. Thus, there is a bijection between the set of $$2$$-Sylow subgroups of such a group and the set of cyclic subgroups of order four.

Inside the symmetric group of degree four
The dihedral group of order eight is the $$2$$-Sylow subgroup inside the symmetric group of degree four. There are three $$2$$-Sylow subgroups, and these correspond to the three cyclic subgroups of order four. Namely:

$$\langle (1,2,3,4) \rangle \leftrightarrow \langle (1,2,3,4), (1,3) \rangle, \langle (1,3,4,2) \rangle \leftrightarrow \langle (1,3,4,2), (1,4) \rangle, \langle (1,4,2,3) \rangle \leftrightarrow \langle (1,4,2,3), (1,2) \rangle$$.

The Sylow subgroup in each case is the normalizer of the corresponding cyclic maximal subgroup.

Inside the projective special linear group $$PSL(3,2)$$
The dihedral group of order eight is a 2-Sylow subgroup inside this group as well.

Comparison with cyclic maximal subgroup in other dihedral groups
In any dihedral group, given by:

$$D_{2n} = \langle a,x \mid a^n = x^2 = e, xax = a^{-1} \rangle$$.

For $$n \ge 3$$, the cyclic subgroup $$\langle a \rangle$$ is a characteristic subgroup. However, for all $$n \ne 4$$, the cyclic subgroup can be characterized as the centralizer of commutator subgroup. But, in the case $$n = 4$$ (which is the one we're considering here), the cyclic maximal subgroup is not the centralizer of commutator subgroup.

Here are some of the salient differences between the cyclic maximal subgroup in the dihedral group of order eight and the cyclic maximal subgroup in the dihedral groups of order $$2^n, n \ge 4$$:


 * In the dihedral group of order eight, the cyclic maximal subgroup is not the centralizer of derived subgroup. In dihedral groups of order $$2^n, n \ge 4$$, it is the centralizer of derived subgroup.
 * In the dihedral group of order eight, the cyclic maximal subgroup is an abelian subgroup of maximum order, but it is not the only one. In dihedral groups of order $$2^n, n \ge 4$$, it is the unique abelian subgroup of maximum order. In particular, in the higher cases, the cyclic maximal subgroup equals the join of abelian subgroups of maximum order (called the Thompson subgroup) and hence also equals the ZJ-subgroup (the center of the join of abelian subgroups of maximum order).
 * In the dihedral group of order eight, the cyclic maximal subgroup is one out of four different centrally large subgroups. In dihedral groups of order $$2^n, n \ge 4$$, it is the unique centrally large subgroup.
 * In the dihedral group of order eight, the cyclic maximal subgroup is one of four subgroups with abelianization of maximum order. In dihedral groups of order $$2^n, n \ge 4$$, it is the unique subgroup with abelianization of maximum order.

Comparison with cyclic maximal subgroup in other semidihedral groups
We define, for $$n \ge 4$$:

$$SD_{2^n} = \langle a,x \mid a^{2^{n-1}} = x^2 = e, xax^{-1} = a^{2^{n-2} - 1} \rangle$$.

Comparison with the $$M_{2^n}$$-groups
We define, for $$n \ge 3$$:

$$M_{2^n} = \langle a,x \mid a^{2^{n-1}} = x^2 = e, xax^{-1} = a^{2^{n-2} + 1} \rangle$$.

Finding this subgroup in a black-box dihedral group
Suppose we know that $$G$$ is a group isomorphic to dihedral group:D8. Then, we can define the subgroup $$H$$ as follows:

H := Filtered(NormalSubgroups(G),K -> IdGroup(K) = [4,1])[1];