Non-central Z4 in M16

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) cyclic group:Z4 and the group is (up to isomorphism) M16 (see subgroup structure of M16).
The subgroup is a normal subgroup and the quotient group is isomorphic to cyclic group:Z4.
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Definition

We consider the group:

$G = M_{16} = \langle a,x \mid a^8 = x^2 = e, xax = a^5 \rangle$

with $e$ denoting the identity element.

This is a group of order 16, with elements:

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

We are interested in the subgroup:

$H := \{ e, a^2x, a^4, a^6x \}$

This subgroup is isomorphic to cyclic group:Z4, with generator $a^2x$. It is a normal subgroup and the quotient group is also isomorphic to the cyclic group:Z4.

Here is the multiplication table for $H$. Note that $H$ is an abelian group so we don't have to worry about left/right issues:

Element/element $e$ $a^2x$ $a^4$ $a^6x$
$e$ $e$ $a^2x$ $a^4$ $a^6x$
$a^2x$ $a^2x$ $a^4$ $a^6x$ $e$
$a^4$ $a^4$ $a^6$ $e$ $a^2$
$a^6x$ $a^6x$ $e$ $a^2x$ $a^4$

Cosets

The subgroup is a normal subgroup and so its left cosets are the same as its right cosets. It has 4 cosets in the whole group:

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

Below is the multiplication for $G/H$. It is a cyclic group of order 4 with generator the coset $\{ a, a^3x, a^5, a^7x \}$. Since it is an abelian group, we do not need to worry about left/right convention for the multiplication table:

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

Subgroup properties

Invariance under automorphisms and endomorphisms

Property Meaning Satisfied? Explanation
normal subgroup invariant under inner automorphisms Yes Follows from being the only non-central cyclic group of order four.
characteristic subgroup invariant under all automorphisms Yes Follows from being the only non-central cyclic group of order four.
fully invariant subgroup invariant under all endomorphisms No The endomorphism $a \mapsto e, x \mapsto x$ (which is a retraction onto $\langle x \rangle$) does not preserve this subgroup.
isomorph-free subgroup no other isomorphic subgroup No center of M16 is another isomorphic subgroup.
isomorph-characteristic subgroup every isomorphic subgroup is characteristic Yes The only isomorphic subgroups are the subgroup itself and center of M16.

GAP implementation

The group and subgroup pair can be constructed as follows:

G := SmallGroup(16,6); H := Filtered(NormalSubgroups(G),x -> Order(x) = 4 and IsCyclic(x) and not(x = Center(G)))[1];

Here is the GAP display:

gap> G := SmallGroup(16,6); H := Filtered(NormalSubgroups(G),x -> Order(x) = 4 and IsCyclic(x) and not(x = Center(G)))[1];
<pc group of size 16 with 4 generators>
Group([ f2*f3, f4 ])

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

gap> Order(G);
16
gap> Order(H);
4
gap> Index(G,H);
4
gap> StructureDescription(H);
"C4"
gap> StructureDescription(G/H);
"C4"
gap> IsNormal(G,H);
true
gap> IsCharacteristicSubgroup(G,H);
true
gap> IsFullinvariant(G,H);
false