Track which structure a map preserves

Try

Reduce an integer modulo five. Adding two integers and then reducing gives the same result as reducing first and adding residues. This compatibility is the defining property of a homomorphism. It preserves the operation, although it may identify many different inputs.

Compare the map from additive integers to integers given by x↦x+1. It is a bijection, but φ(x+y)=x+y+1 differs from φ(x)+φ(y)=x+y+2. A bijection alone is not a group isomorphism.

Predict

For φ:G→K satisfying φ(ab)=φ(a)φ(b), predict what φ(e) must be. Do not assume identity preservation as a separate unexplained rule. Use the equation φ(e)=φ(e)φ(e) and cancellation in the target group to derive it.

Experiment

Consider φ:Z₁₀→Z₁₀ defined additively by φ(x)=2x. List its outputs.

Inputs sharing an output	Output
0,5	0
1,6	2
2,7	4
3,8	6
4,9	8

The kernel is ｛0,5｝; the image is ｛0,2,4,6,8｝. Each fiber is a coset of the kernel. The map is neither injective nor onto its declared codomain, but it is onto its image by definition. State the codomain explicitly before making a surjectivity claim.

Understand

Once φ(e)=e is proved, applying φ to gg⁻¹=e shows φ(g⁻¹)=φ(g)⁻¹. The kernel consists of elements sent to identity. It is a subgroup by the one-step test, and it is normal because φ(gng⁻¹)=φ(g)eφ(g)⁻¹=e for every kernel element n. The image is a subgroup because φ(a)φ(b)⁻¹=φ(ab⁻¹). An image need not be normal in the whole codomain; inclusion of a nonnormal subgroup is a counterexample.

The kernel detects injectivity. If φ is injective, the only preimage of identity is e. Conversely, if φ(a)=φ(b), then b⁻¹a lies in the kernel. A trivial kernel forces a=b. This proof works without finiteness, while a size-only argument would not.

Fibers are kernel cosets. The equality φ(g)=φ(a) is equivalent to φ(a⁻¹g)=e, which means g∈a ker φ. Thus every nonempty fiber has the same size as the kernel when the groups are finite. The next lesson turns this observation into an isomorphism theorem.

Some plausible maps fail to preserve multiplication. Squaring in an abelian group works because (ab)²=a²b² after commuting the middle factors. In S₃, let a=(1 2),b=(2 3). Then a²b²=e, while ab is a 3-cycle and (ab)² is not identity. The missing hypothesis is visible in one explicit pair.

Apply

For a proposed map, check that it is well-defined on the stated domain before checking the operation. For residue groups, different integer representatives must produce the same output residue. Then calculate kernel and image, and only afterward decide injectivity and surjectivity.

Write a proof that a homomorphism sends an element of finite order n to an element whose order divides n. Explain why the order can decrease when the kernel is nontrivial. A structure-preserving map need not preserve every distinction between inputs.

Continue

The first isomorphism theorem identifies precisely the group obtained after collapsing kernel fibers. It replaces a potentially many-to-one map by an isomorphism from a quotient onto the image.