Let

$$ Q_{4n} = \langle a,b:a^{2n}=1,\ b^2=a^n,\ b^{-1}ab=a^{-1}\rangle, \qquad n\geq3\text{ odd}, $$

and write

$$ z=a^n. $$

This is the unique involution in \(Q_{4n}\). Quotienting by it gives the dihedral group

$$ \pi:Q_{4n}\longrightarrow D_{2n}. $$

Start with an inverse-closed connection set \(\bar S\subset D_{2n}\setminus\{1\}\). There are two especially clean ways to lift it:

$$ S=\pi^{-1}(\bar S) \qquad\text{or}\qquad S=\pi^{-1}(\bar S)\cup\{z\}. \tag{1} $$

If \(\operatorname{Cay}(D_{2n},\bar S)\) is twin-free and is a CI graph on \(D_{2n}\), then both graphs in (1) are CI graphs on \(Q_{4n}\).

The cute part is that the quotient map is not supplied to an isomorphism. The graph has to recover it by itself.

The two lifts are blow-ups#

Put

$$ \bar\Gamma=\operatorname{Cay}(D_{2n},\bar S). $$

The first lift in (1) replaces every vertex of \(\bar\Gamma\) by two nonadjacent vertices. The second replaces every vertex by an edge:

$$ \operatorname{Cay}(Q_{4n},\pi^{-1}(\bar S)) = \bar\Gamma[\overline{K_2}], $$

$$ \operatorname{Cay}(Q_{4n},\pi^{-1}(\bar S)\cup\{z\}) = \bar\Gamma[K_2]. \tag{2} $$

The two vertices above one quotient vertex are

$$ \{g,zg\}. \tag{3} $$

They are twins: outside their own pair they have exactly the same neighborhood. In the second lift they are adjacent twins; in the first they are nonadjacent twins.

Because \(\bar\Gamma\) is twin-free, there are no accidental mergers between different quotient vertices. Thus the pairs in (3) are exactly the maximal twin classes of the lifted graph.

That makes the central quotient visible without labels:

$$ \boxed{ \text{the graph's intrinsic two-point twin classes are the fibers of }\pi. } \tag{4} $$

Every Cayley target has the same pairs#

Suppose another Cayley graph on the same abstract quaternion group is isomorphic to the lift:

$$ \operatorname{Cay}(Q_{4n},S) \cong \operatorname{Cay}(Q_{4n},T). \tag{5} $$

The target graph has the same intrinsic partition into \(2n\) twin classes of size two. Its right-regular \(Q_{4n}\) action permutes these classes transitively.

A class stabilizer therefore has order

$$ \frac{|Q_{4n}|}{2n}=2. $$

But \(Q_{4n}\) has only one subgroup of order two:

$$ \langle z\rangle. $$

So the target's twin classes are also exactly the central pairs \(\{g,zg\}\). Looking at the neighborhood of the identity now forces \(T\) to have one of the same two uniform forms:

$$ T=\pi^{-1}(\bar T) \qquad\text{or}\qquad T=\pi^{-1}(\bar T)\cup\{z\}. \tag{6} $$

An arbitrary isomorphic Cayley target could have been nonuniform over the central fibers. The twin partition rules that out.

The isomorphism in (5) descends to

$$ \operatorname{Cay}(D_{2n},\bar S) \cong \operatorname{Cay}(D_{2n},\bar T). \tag{7} $$

It also remembers whether each twin pair contains an edge, so the two choices in (1) and (6) agree.

The quotient automorphism always lifts#

Since \(\bar\Gamma\) is a CI graph, (7) comes from an automorphism of the dihedral group. Every such automorphism has the form

$$ r\longmapsto r^u, \qquad s\longmapsto r^v s, \tag{8} $$

where \(u\) is a unit modulo \(n\).

Exactly one of \(u\) and \(u+n\) is an odd unit modulo \(2n\). Call that choice \(u'\). Then

$$ a\longmapsto a^{u'}, \qquad b\longmapsto a^v b \tag{9} $$

is an automorphism of \(Q_{4n}\). It lifts (8) and fixes \(z\).

Therefore it carries \(S\) to \(T\). This proves the CI claim.

I like how little machinery survives in the final proof. The first approach looked like a regular-subgroup cohomology problem. The graph then noticed its own twin pairs and made the quotient unavoidable.

The square-free family#

The generalized-dihedral classification gives a large immediate family. When \(n\) is odd and square-free, \(D_{2n}\) is a CI group. Hence every central-uniform lift in (1) with a twin-free quotient is a CI graph on \(Q_{4n}\).

For example, every such lift from \(D_{30}\) to \(Q_{60}\) is CI.

This does not say that \(Q_{60}\) is a CI group. It removes one entire region of its connection-set space:

$$ \boxed{ \text{a residual }Q_{60}\text{ defect must either have quotient twins or be nonuniform over some central pair.} } \tag{10} $$

That is the exact boundary. Twin-freeness is load-bearing because it makes the central fibers intrinsic. If quotient vertices are already twins, several central pairs can merge into a larger twin class and an isomorphism need not preserve the quotient. Nonuniform lifts escape before the blow-up description in (2) even begins.

The verifier checks all \(120\) automorphisms of \(D_{30}\) lift explicitly, exhausts the central-uniform cases for \(Q_{12}\) and \(Q_{20}\), and replays the \(Q_{60}\) extension-cocycle support. The theorem itself is all-order: the finite calculations are there to be rude to the proof, not to replace it.

Notebook references: R-0761, R-0891