Make multiplication independent of representatives

Try

In integers modulo eight, let N=｛0,4｝. Its cosets are ｛0,4｝,｛1,5｝,｛2,6｝,｛3,7｝. Add any representative from the first nonzero coset to any representative from the next one. The answer always belongs to ｛3,7｝. That consistency is what makes a quotient operation meaningful.

In a nonabelian group, the same consistency can fail. A quotient group requires more than a partition into equal-sized subsets.

Predict

Suppose a is replaced by an and b by bm, with n,m∈N. The product becomes anbm. Predict which element must stay in N to compare this with ab. Moving n past b requires conjugation, not commutativity of all group elements.

Experiment

Track the representative change without exchanging factors illegally.

Expression	Justification
(an)(bm)	New representatives
ab(b⁻¹nb)m	Insert bb⁻¹ and reassociate
abn′m with n′∈N	Conjugation invariance
Same coset as ab	Closure in N

Thus normality supplies exactly the missing step. For the residue example, the quotient has four elements and is cyclic, generated by 1+N. Although each quotient element is a two-element subset, its powers count returns to the identity subset N.

Understand

A subgroup N is normal in G when gNg⁻¹=N for every g∈G. Equivalently gN=Ng for every g. Multiplying either set equality by g or g⁻¹ proves the equivalence. Normality does not mean every n commutes with every g; it means conjugation preserves the subgroup as a set.

For normal N, define (aN)(bN)=abN. The representative calculation proves well-definedness. Associativity follows from G, the identity is N, and the inverse of aN is a⁻¹N. The resulting group is G/N. The quotient map g↦gN preserves multiplication and has kernel N.

For a concrete failure, take H=｛e,(1 2)｝ in S₃ and b=(1 2 3). The elements e and (1 2) represent the same coset H. Their products with b would need to determine the same output coset. But bH and (1 2)bH are different: equality would require b⁻¹(1 2)b=(1 3)∈H, which is false. One inconsistent representative choice is enough to defeat the proposed operation.

Every index-two subgroup is normal. If g belongs to H, both gH and Hg equal H. Otherwise both are the complement of H, because each left and right partition has exactly two parts. This proves equality for every g. In particular, rotations form a normal subgroup in Dₙ, and the quotient records only whether a symmetry preserves or reverses orientation.

Apply

Use the quotient of residues modulo eight by ｛0,4｝ to build a four-by-four addition table. Check that it matches addition modulo four. Then explain the map from the original group onto that smaller group and identify exactly which elements become identity.

Repair the statement “all subgroups of a group are normal because left and right cosets have the same size.” Equal cardinality is not equality of subsets. Supply the S₃ conjugation counterexample and identify the stronger condition actually needed.

Transfer: when does a map survive a quotient?

A quotient identifies elements. A proposed later measurement must give the same answer on every representative of an identified class. Let N be normal in G and let φ:G→H be a group homomorphism. There is a homomorphism φ̄:G/N→H satisfying φ̄(gN)=φ(g) if and only if N⊆ker φ.

Necessity. Every n∈N represents the identity coset N, so φ(n) must equal the identity of H. Sufficiency. If gN=hN, then h⁻¹g∈N. Applying φ gives φ(h)⁻¹φ(g)=e, hence φ(g)=φ(h). Thus the proposed rule is independent of representatives. It preserves multiplication because φ does, and it is unique because every coset has a representative g. This argument states exactly what must be checked; a diagram of arrows alone does not prove well-definedness.

The image of φ̄ is im φ, while its kernel is the subgroup (ker φ)/N. Therefore a descended map need not be injective. It is injective precisely when N=ker φ. It need not reach every element of H either: descent does not enlarge the original image.

Worked changed case. In additive G=Z/12, take N generated by 4, namely the three classes 0,4,8. The map φ([x])=[3x] kills N and descends to a map from the four-element quotient. Its kernel is exactly N, so the descended map is injective, with image the four classes 0,3,6,9. It is not onto all twelve classes. In contrast ψ([x])=[2x] does not descend: 0 and 4 lie in the same coset but map to 0 and 8.

Predict, then inspect the certificate

Changed case A map that descendsA map that does not descend

In additive Z/12, N=⟨4⟩. Does φ([x])=[3x] in Z/12 descend to G/N?

Your prediction Choose a predictionYesNo Check reasoning Reset prediction

Two authored cases are practice, not a proof of the general theorem. Explain which hypothesis changed. Responses are temporary and do not change saved progress.

Changed task. Keep G and N, but use θ([x])=[6x]. Decide whether it descends, whether the induced map is injective and whether it is onto H=Z/12. Give a kernel and image certificate.

Feedback

It descends since 6·4=0 modulo 12. Its original kernel is all six even classes, containing N properly. The induced kernel has two cosets, so the four-element quotient map is not injective. Its image is only 0 and 6, so it is not onto H. Distinguish the original kernel size six from the quotient kernel size two.

Continue

Homomorphisms connect these quotient constructions to maps between groups. Their kernels explain which elements are collapsed, while their images explain which target elements are actually reached.