D8 in SD16

This article is about a particular subgroup in a group, up to equivalence of subgroups (i.e., an isomorphism of groups that induces the corresponding isomorphism of subgroups). The subgroup is (up to isomorphism) dihedral group:D8 and the group is (up to isomorphism) semidihedral group:SD16 (see subgroup structure of semidihedral group:SD16).
The subgroup is a normal subgroup and the quotient group is isomorphic to cyclic group:Z2.
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Definition

Here, $G$ is the semidihedral group:SD16, the semidihedral group of order sixteen (and hence, degree eight). We use here the presentation:

$G := \langle a,x \mid a^8 = x^2 = e, xax = a^3 \rangle$

$G$ has 16 elements:

$\! \{ e,a,a^2,a^3,a^4,a^5,a^6,a^7,x,ax,a^2x,a^3x,a^4x,a^5x,a^6x,a^7x \}$

The subgroup $H$ of interest is the subgroup $\langle a^2,x \rangle$. It is dihedral of order 8 and is given by:

$\! \{ e,a^2,a^4,a^6,x,a^2x,a^4x,a^6x \}$

The multiplication table of $H$ is as follows:

Element	$\! e$	$\! a^2$	$\! a^4$	$\! a^6$	$\! x$	$\! a^2x$	$\! a^4x$	$\! a^6x$
$\! e$	$\! e$	$\! a^2$	$\! a^4$	$\! a^6$	$\! x$	$\! a^2x$	$\! a^4x$	$\! a^6x$
$\! a^2$	$\! a^2$	$\! a^4$	$\! a^6$	$\! e$	$\! a^2x$	$\! a^4x$	$\! a^6x$	$\! x$
$\! a^4$	$\! a^4$	$\! a^6$	$\! e$	$\! a^2$	$\! a^4x$	$\! a^6x$	$\! x$	$\! a^2x$
$\! a^6$	$\! a^6$	$\! e$	$\! a^2$	$\! a^4$	$\! a^6x$	$\! x$	$\! a^2x$	$\! a^4x$
$\! x$	$\! x$	$\! a^6x$	$\! a^4x$	$\! a^2x$	$\! e$	$\! a^6$	$\! a^4$	$\! a^2$
$\! a^2x$	$\! a^2x$	$\! x$	$\! a^6x$	$\! a^4x$	$\! a^2$	$\! e$	$\! a^6$	$\! a^4$
$\! a^4x$	$\! a^4x$	$\! a^2x$	$\! x$	$\! a^6x$	$\! a^4$	$\! a^2$	$\! e$	$\! a^6$
$\! a^6x$	$\! a^6x$	$\! a^4x$	$\! a^2x$	$\! x$	$\! a^6$	$\! a^4$	$\! a^2$	$\! e$

Cosets

The subgroup has index two and is hence a normal subgroup (See index two implies normal). Its left cosets coincide with its right cosets. There are two cosets:

$\! H = \{ e,a^2,a^4,a^6,x,a^2x,a^4x,a^6x \}, G \setminus H = \{ a, a^3, a^5, a^7, ax, a^3x, a^5x, a^7x \}$

Arithmetic functions

Function	Value	Explanation
order of whole group	16
order of subgroup	8
index of subgroup	2
size of conjugacy class (=index of normalizer)	1
number of conjugacy classes in automorphism class	1
size of automorphism class of subgroup	1

Subgroup-defining functions

Subgroup-defining function	Meaning in general	Why it takes this value
first omega subgroup	subgroup generated by elements of order two	The elements of order two are $\{ a^4, x, a^2x, a^4x, a^6 x\}$. These generate this subgroup.
join of abelian subgroups of maximum rank	subgroup generated by the abelian subgroups that have maximum rank	There are no rank 3 abelian subgroups, and the rank 2 abelian subgroups are $\langle a^4, x \rangle$ and $\langle a^4, a^2x \rangle$ -- both Klein four-groups (see V4 in SD16). Together, they generate the dihedral group.
join of elementary abelian subgroups of maximum order	subgroup generated by the elementary abelian subgroups of maximum order	There are no elementary abelian subgroups of order 8, and the elementary abelian subgroups of order 4 are $\langle a^4, x \rangle$ and $\langle a^4, a^2x \rangle$ -- both Klein four-groups (see V4 in SD16). Together, they generate the dihedral group.

Subgroup properties

Invariance under automorphisms and endomorphisms

Property	Meaning	Satisfied?	Explanation
normal subgroup	invariant under inner automorphisms	Yes	index two implies normal
characteristic subgroup	invariant under all automorphisms	Yes	On account of being the first omega subgroup, also on account of being an isomorph-free subgroup.
fully invariant subgroup	invariant under all endomorphisms	Yes	On account of being the first agemo subgroup.
isomorph-free subgroup	no other isomorphic subgroup	Yes	The other two subgroups of order 8 are Z8 in D16 and Q8 in D16, both non-isomorphic to it. Also follows from being the first omega subgroup.
homomorph-containing subgroup	contains every homomorphic image	Yes	On account of being an omega subgroup.
image-closed characteristic subgroup	under any surjective homomorphism from $G$ to a group $K$, the image of $H$ is characteristic in $K$.	No	If we consider $K = G/Z(G)$ and take the quotient map, then the image of $H$ in there looks like one of the Klein four-subgroups of dihedral group:D8, which is not characteristic.
image-closed fully invariant subgroup	under any surjective homomorphism from $G$ to a group $K$, the image of $H$ is fully invariant in $K$.	No	Follows from not being image-closed characteristic.
verbal subgroup	generated by set of words	No	Follows from not being image-closed fully invariant.

GAP implementation

The group and subgroup pair can be constructed using GAP as follows:

G := SmallGroup(16,8); H := Filtered(Subgroups(G), x -> Order(x) = 8 and IsDihedralGroup(x))[1];

The GAP display is as follows:

gap> G := SmallGroup(16,8); H := Filtered(Subgroups(G), x -> Order(x) = 8 and IsDihedralGroup(x))[1];
<pc group of size 16 with 4 generators>
Group([ f4, f3, f2 ])

Here is GAP code to verify some of the assertions on this page:

gap> Order(G);
16
gap> Order(H);
8
gap> Index(G,H);
2
gap> StructureDescription(H);
"D8"
gap> StructureDescription(G/H);
"C2"
gap> IsNormal(G,H);
true
gap> IsCharacteristicSubgroup(G,H);
true
gap> IsFullinvariant(G,H);
true