# Supergroups of quaternion group

## Contents

View supergroups of particular groups | View other specific information about quaternion group

Note that unlike the discussion of the subgroup structure of quaternion group, this discussion is necessarily not comprehensive, because there are infinitely many groups containing the dihedral group of order eight. However, we provide a comprehensive discussion of all the groups of order sixteen containing this as a subgroup.

## Analysis of groups containing this as a subgroup of index two

### The general procedure

Note first that any subgroup of index two is normal, so the groups we are interested in classifying have the dihedral group of order eight as a normal subgroup of index two.

We can use cohomology theory to begin this analysis. Specifically, we are interested in groups of order sixteen where the quaternion group of order eight is the normal subgroup and the quotient group is the cyclic group of order two. We denote by $N$ the quaternion group of order eight and $Q$ the quotient group, which is cyclic of order two.

The classification proceeds in three steps:

• Determine the set of possible homomorphisms $Q \to \operatorname{Out}(N)$.
• For each such homomorphism, determine whether an extension exists.
• If an extension exists, classify the extensions using the second cohomology group $H^2(Q,Z(N))$ corresponding to the induced action on $Z(N)$ from the homomorphism to $\operatorname{Out}(N)$.

### The classification $\operatorname{Out}(N)$ is isomorphic to the symmetric group of degree three. This has three subgroups of order two, but these are conjugate in $\operatorname{Out}(N)$. Thus, up to equivalence under automorphisms, there is a unique subgroup of order two in $\operatorname{Out}(N)$.

Thus, up to equivalence under automorphisms, there are two homomorphisms from $Q$ to $\operatorname{Out}(N)$: the trivial map and the map that goes isomorphically to one of the subgroups of order two.

For the trivial map, $H^2(Q,Z(N))$ corresponds to the trivial action and has order two, so the two supergroups are:

For the isomorphism to one of the subgroups of order two, $H^2(Q,Z(N))$ corresponds to the trivial action and has order two, so the two supergroups are: