Center of quaternion group

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:Z2 and the group is (up to isomorphism) quaternion group (see subgroup structure of quaternion group).
The subgroup is a normal subgroup and the quotient group is isomorphic to Klein four-group.
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Definition

This article is about the quaternion group:

$G := \{ 1, -1, i, -i, j, -j, k,-k \}$

with multiplication table:

In the table below, the row element is multiplied on the left and the column element on the right.

Element	$\! 1$	$\! -1$	$\! i$	$\! -i$	$\! j$	$\! -j$	$\! k$	$\! -k$
$\! 1$	$\! 1$	$\! -1$	$\! i$	$\! -i$	$\! j$	$\! -j$	$\! k$	$\! -k$
$\! -1$	$\! -1$	$\! 1$	$\! -i$	$\! i$	$\! -j$	$\! j$	$\! -k$	$\! k$
$\! i$	$\! i$	$\! -i$	$\! -1$	$\! 1$	$\! k$	$\! -k$	$\! -j$	$\! j$
$\! -i$	$\! -i$	$\! i$	$\! 1$	$\! -1$	$\! -k$	$\! k$	$\! j$	$\! -j$
$\! j$	$\! j$	$\! -j$	$\! -k$	$\! k$	$\! -1$	$\! 1$	$\! i$	$\! -i$
$\! -j$	$\! -j$	$\! j$	$\! k$	$\! -k$	$\! 1$	$\! -1$	$\! -i$	$\! i$
$\! k$	$\! k$	$\! -k$	$\! j$	$\! -j$	$\! -i$	$\! i$	$\! -1$	$\! 1$
$\! -k$	$\! -k$	$\! k$	$\! -j$	$\! j$	$\! i$	$\! -i$	$\! 1$	$\! -1$

We are interested in the subgroup:

$H = \{ 1, -1 \}$

with multiplication table:

Element/element	1	-1
1	1	-1
-1	-1	1

Cosets

The subgroup is a normal subgroup, so its left cosets coincide with its right cosets. There are four cosets, given as follows:

$\! \{ 1,-1\}, \{ i,-i \}, \{ j,-j \}, \{ k,-k \}$

The quotient group is isomorphic to a Klein four-group, and its multiplication table is as given below:

Element/element	$\! \{ 1, -1 \}$	$\! \{ i, -i \}$	$\! \{ j, -j \}$	$\! \{k -k \}$
$\! \{ 1, -1 \}$	$\! \{ 1, -1 \}$	$\! \{ i, -i \}$	$\! \{ j, -j \}$	$\! \{ k, -k \}$
$\! \{ i, -i \}$	$\! \{ i, -i \}$	$\! \{ 1, -1 \}$	$\! \{ k, -k \}$	$\! \{ j, -j \}$
$\! \{ j, -j \}$	$\! \{ j, -j \}$	$\! \{ k, -k \}$	$\! \{ 1, -1 \}$	$\! \{ i, -i \}$
$\! \{ k, -k \}$	$\! \{ k, -k \}$	$\! \{ j, -j \}$	$\! \{ i, -i \}$	$\! \{ 1, -1 \}$

Subgroup-defining functions

Subgroup-defining function	What it means in general	Why it takes this value	Corresponding quotient-defining function	GAP verification (set G := SmallGroup(8,4); H := Center(G);) -- see #GAP implementation
center	set of elements that commute with every element		inner automorphism group	Definitional
derived subgroup	subgroup generated by commutators of elements		abelianization	H = DerivedSubgroup(G); using DerivedSubgroup
Frattini subgroup	intersection of all maximal subgroups	It is the intersection of the three maximal subgroups, which are $\langle i \rangle$, $\langle j \rangle$, and $\langle k \rangle$.	Frattini quotient	H = FrattiniSubgroup(G); using FrattiniSubgroup
socle	join of all minimal normal subgroups	It is the unique minimal normal subgroup	?	H = Socle(G); using Socle
First omega subgroup	subgroup generated by elements of order $p$, in this case two		?
First agemo subgroup	subgroup generated by $p^{th}$ powers, in this case squares		?	H = Agemo(G,2,1); using Agemo

Subgroup properties

Invariance under automorphisms and endomorphisms

Property	Meaning	Satisfied?	Explanation
normal subgroup	invariant under inner automorphisms	Yes	center is normal, derived subgroup is normal
characteristic subgroup	invariant under all automorphisms	Yes	center is characteristic, derived subgroup is characteristic
fully invariant subgroup	invariant under all endomorphisms	Yes	derived subgroup is fully invariant
isomorph-free subgroup, isomorph-containing subgroup	no other isomorphic subgroups	Yes	only subgroup of order two
homomorph-containing subgroup	contains every homomorphic image	Yes	on account of its non-identity elements being precisely the elements that are of order two.
variety-containing subgroup		Yes
verbal subgroup	generated by set of words	Yes	derived subgroup is verbal, agemo subgroups are verbal

Centrality and related properties

Property	Meaning	Satisfied?	Explanation
central subgroup	contained in the center	Yes
central factor	product with centralizer is whole group	Yes

Cohomology interpretation

We can think of $G$ as an extension with abelian normal subgroup $H$ and quotient group $G/H$. Since $H$ is in fact the center, the action of the quotient group on the normal subgroup is the trivial group action. We can thus study $G$ as an extension group arising from a cohomology class for the trivial group action of $G/H$ (which is a Klein four-group) on $H$ (which is cyclic group:Z4).

For more, see second cohomology group for trivial group action of V4 on Z4.