# 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)));

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)));
<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