Try
A group with eight elements need not look like eight rotations. Start with a square: let a rotate its four vertices by one step and let b reflect them. The eight distinct symmetries are aⁱ and aⁱb for 0≤i≤3. Here a⁴=1, b²=1 and bab⁻¹=a⁻¹. We write this group as D₄, of order 8; some books instead call it D₈.
Calculate (ab)² using the conjugation relation, not by commuting a past b. Then predict what changes if the reflection-like generator instead satisfies b²=a². Will the relations still describe a group of eight elements, and will the elements outside the rotation subgroup still have order two?
Recall dihedral normal forms, finite abelian structure and semidirect products. Our goal is an exhaustive classification, so we must prove that the proposed constructions exist, that no other groups occur, and that the resulting groups are pairwise nonisomorphic.
Predict
Suppose G is nonabelian and has order 8. Lagrange’s theorem allows element orders 1,2,4,8. Explain why an element of order 8 is impossible. Next suppose every element satisfies x²=1. The inverse-of-a-product identity then gives xy=(xy)⁻¹=y⁻¹x⁻¹=yx. Thus this alternative is abelian too. Consequently a nonabelian G must contain an element a of order 4.
The subgroup H=⟨a⟩ has index two and is normal: for an element outside H, both its left and right cosets are the complement of H. Choose b outside H. Every element has a unique form aⁱbʲ with 0≤i≤3 and j=0 or 1. Before reading further, predict the only two possibilities for the automorphism of H induced by conjugation by b.
Experiment
An automorphism sends the generator a to a or a³. If bab⁻¹=a, then a and b commute; since they generate G, G is abelian. In our nonabelian case conjugation must invert a. Also b² lies in H because G/H has order two. Write b²=aˢ. Conjugation by b fixes b², but sends aˢ to a⁻ˢ. Hence a²ˢ=1: s is 0 or 2 modulo 4.
Use these facts to fill the last two columns independently, then compare your calculations with the table.
| Relation for b² | Square of aⁱb | Nonidentity elements of order two | Elements of order four |
|---|
| b²=1 | 1 | a²,b,ab,a²b,a³b: five | a,a³: two |
| b²=a² | a² | a²: one | a,a³,b,ab,a²b,a³b: six |
Indeed, (aⁱb)²=aⁱa⁻ⁱb²=b². In both cases the exact multiplication rule is
(aⁱbʲ)(aᵏbˡ) = a^(i + (−1)ʲ k + s j l) b^((j+l) mod 2),
where the exponent of a is reduced modulo 4, and s is 0 or 2. In the displayed rule “mod 2” applies to the exponent j+l of b only. The extra term s j l records the carry b² when j=l=1. This is a multiplication recipe derived inside a group, not yet a proof that an arbitrary recipe is associative.
Understand
Construct both cases and justify exhaustion. For s=0, the square symmetries from Try give a group of eight elements with the required relations. For s=2, work with invertible complex matrices
A=diag(i,−i), B with rows (0,1),(−1,0).
Matrix multiplication gives A²=B²=−I, A⁴=I and BAB⁻¹=A⁻¹. The eight matrices Aʲ and AʲB, 0≤j≤3, are distinct: the first four are diagonal with distinct first entries, the other four have zero diagonal and distinct upper-right entries. The multiplication rule above proves closure. It also supplies inverses (or use finite closure inside a group). Associativity is inherited from matrix multiplication. This group is Q₈, the quaternion group; its elements are commonly written ±1,±i,±j,±k.
To ensure relations do not hide extra possibilities, start with the abstract presentation a⁴=1, bab⁻¹=a⁻¹, b²=aˢ. The relations move every a to the left and reduce every word to one of the eight normal forms, including inverses. Thus the presented group has at most eight elements. The corresponding concrete construction has eight distinct such elements, so the presentation has exactly eight. Any nonabelian group G of order 8 considered above is generated by a,b satisfying one presentation. The induced surjection onto G is therefore an isomorphism. This proves exhaustion, rather than just finding two examples.
Include and separate the abelian cases. The finite abelian classification gives C₈, C₄×C₂ and C₂×C₂×C₂: these correspond to the three partitions of 3 into positive exponents of 2. Their maximum element orders are respectively 8,4 and 2, so they are pairwise nonisomorphic. Both D₄ and Q₈ are nonabelian, since conjugation inverts an element of order four. They are not isomorphic to each other because the number of elements of order two is five versus one. These arguments establish exactly five isomorphism types.
The following decision table is a textual classification tree. Follow one row from its first test to its conclusion; each leaf is backed by the preceding existence and exhaustion proofs. For an abelian group whose maximum element order is four, predict its leaf before looking at the last column.
| First branch | Next test | Forced structure | Classification leaf |
|---|
| Abelian | Has an element of order 8 | One cyclic factor | C₈ |
| Abelian | Maximum element order 4 | Factors of orders 4 and 2 | C₄×C₂ |
| Abelian | All nonidentity orders 2 | Three factors of order 2 | C₂³ |
| Nonabelian | Outside square b²=1 | Inversion action with outside involution | D₄ |
| Nonabelian | Outside square b²=a² | Inversion action with outside elements of order 4 | Q₈ |
There are no missing branches: Lagrange restricts element orders, the abelian theorem handles the first three rows, and the forced order-four subgroup and conjugation argument restrict the nonabelian rows to exactly the two displayed squares.
Identify which extension splits. In Q₈ the normal cyclic subgroup ⟨a⟩ has order four and quotient C₂. A splitting would send the nonidentity element of C₂ to an element outside ⟨a⟩ whose square is 1. Every such element instead squares to a²≠1. Thus this extension does not split. In D₄, b is an outside involution, giving the complement ⟨b⟩ and the semidirect product C₄⋊C₂ with inversion action.
This does not mean that every split extension of C₄ by C₂ is dihedral: the trivial action gives C₄×C₂, while inversion gives D₄. Nor does “nonsplit” alone characterize Q₈: C₈ has a cyclic subgroup of order four and quotient C₂ with no outside involution, so that extension is also nonsplit. These are statements about specified kernels and quotients. C₂³, for example, splits as C₂²×C₂, but has no cyclic subgroup of order four. The distinction complements the split-extension criteria; it does not assume that all finite extensions split.
Apply
A proposed nonabelian group has a of order four, an outside element b, inversion conjugation and b²=a. Diagnose the contradiction without enumerating a multiplication table. Next, in D₄ use K=⟨a²,b⟩ instead of ⟨a⟩: is K normal, and does the quotient extension split? Explain why this changes the description without changing the group.
Check the two repairs
Conjugating b²=a by b fixes its left side but inverts its right side. Thus a=a⁻¹, contradicting order four. Odd values of s are impossible under the declared hypotheses.
K consists of 1,a²,b,a²b and is C₂×C₂. Its index is two, so it is normal. The involution ab lies outside K and generates a complement; the two subgroups intersect only in 1, and their product has eight elements. Therefore D₄ also has a split description K⋊⟨ab⟩. The action is nontrivial: conjugation by ab fixes a² and sends b to a²b. A semidirect-product description records the chosen normal subgroup and action, not just an abstract group name.
Continue
Write an exhaustion checklist for this proof: first separate abelian and nonabelian groups; next force an order-four element; then determine conjugation and the outside square; finally construct and distinguish every surviving case. Return to small-order classification and identify which steps use the particular number 8. Counting candidate presentations without proving their consistency or exhausting possible groups would not complete a classification.
