# Cyclic maximal subgroups 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:Z4 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 cyclic group:Z2.
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The quaternion group is defined by the following presentation: $\langle i,j,k \mid i^2 = j^2 = k^2 = ijk \rangle$

We use the symbol $1$ for the identity element and the symbol $-1$ for $i^2$. $i^3,j^3,k^3$ are denoted by $-i,-j,-k$ respectively.

It has the following multiplication table:

Element $1$ $-1$ $i$ $-i$ $j$ $-j$ $k$ $-k$ $1$ $1$ $-1$ $i$ $-i$ $j$ $-j$ $k$ $-k$ $-1$ $-1$ $1$ $-i$ $i$ $-j$ $j$ $-k$ $k$ $i$ $i$ $-i$ $-1$ $1$ $k$ $-k$ $-j$ $j$ $-i$ $-i$ $i$ $1$ $-1$ $-k$ $k$ $j$ $-j$ $j$ $j$ $-j$ $-k$ $k$ $-1$ $1$ $i$ $-i$ $-j$ $-j$ $j$ $k$ $-k$ $1$ $-1$ $-i$ $i$ $k$ $k$ $-k$ $j$ $-j$ $-i$ $i$ $-1$ $1$ $-k$ $-k$ $k$ $-j$ $j$ $i$ $-i$ $1$ $-1$

The subgroups that we are interested in studying are the three four-element subgroups of this group, namely the subgroups: $\{ 1,-1,i,-i \}, \qquad \{ 1,-1,j,-j \}, \qquad \{1,-1,k,-k \}$.

## Invariance under automorphisms and endomorphisms: properties satisfied

### Normal subgroup

Further information: normal subgroup, normal subgroup of group of prime power order

All three of these subgroups are normal. The quotient group in each case is a cyclic group of order two.

### Retraction-invariant subgroup

Further information: retraction-invariant subgroup,retraction-invariant normal subgroup

All three of these subgroups are invariant under all retractions of the whole group. This is because of the trivial reason that the group has no retractions under that the identity map and the trivial homomorphism.

## Invariance under automorphisms and endomorphisms: properties not satisfied

### Characteristic subgroup

Further information: characteristic subgroup

None of these three subgroups is characteristic. In fact, there is an automorphism cyclically permuting $i,j,k$ that permutes these subgroups cyclically.

### Coprime automorphism-invariant and cofactorial automorphism-invariant

Further information: coprime automorphism-invariant subgroup, cofactorial automorphism-invariant subgroup

None of these three subgroups is invariant either under the automorphisms whose order is a power of two or under the automorphisms whose order is relatively prime to two. Specifically:

• There is an automorphism of order three cyclically permuting $i,j,k$.
• There is an automorphism of order two that sends $i$ to $-j$, $j$ to $-i$, and $k$ to $-k$. This has order two and it interchanges two of the three subgroups while sending the third subgroup to itself.

## Resemblance notions

All three subgroups are isomorph-automorphic subgroups. In fact, they are order-automorphic subgroups.

## Generic maximality notions

All three subgroups are maximal subgroups of the group of prime power order. Thus, they satisfy all these properties: maximal normal subgroup, maximal subgroup, subgroup of index two, order-normal subgroup, isomorph-normal subgroup, maximal subgroup of finite nilpotent group.

## Maximality notions related to abelianness

### Abelian subgroups of maximum order

All three subgroups are abelian subgroups of maximum order, and they are the only ones. Also, they are maximal among abelian subgroups and maximal among abelian normal subgroups.

### Abelian subgroups of maximum rank

All three subgroups are abelian subgroups of maximum rank. However, they are not the only ones -- so is the center, which has order two.

## GAP implementation

### Finding these subgroups in a black-box quaternion group

Suppose $G$ is a group we know to be abstractly isomorphic to the quaternion group. Then, we can define a three-element list of the subgroups of $G$ using the MaximalSubgroups function as:

L := MaximalSubgroups(G);

Alternatively, we can use the NormalSubgroups function:

L := Filtered(NormalSubgroups(G),H -> Order(H) = 4);

The individual members are accessed as $L, L, L$ respectively.