Replace a map by a quotient and an embedding

Try

Return to multiplication by two on additive residues modulo ten. Its kernel is ｛0,5｝, and its image is the five even residues. The quotient has five cosets. Match each coset to its common image and predict whether this matching preserves addition and is one-to-one.

The purpose of an isomorphism theorem is not merely to compare cardinalities. It produces a specific map and proves that the map preserves the operation.

Predict

Given φ:G→K with kernel N, define a candidate map gN↦φ(g). Which must be proved first: that the formula is independent of the representative, or that it is injective? A formula that assigns different answers to the same coset is not a function at all, so representative independence comes first.

Experiment

Trace the proof obligations for the induced map.

Obligation	Reason
Well-defined	Equal cosets differ by a kernel element
Product preservation	φ(gh)=φ(g)φ(h)
Onto image	Every image element has a preimage
Injective	Equal outputs force equal kernel cosets

In the residue example, ｛1,6｝ maps to two and ｛3,8｝ maps to six. Adding the cosets gives ｛4,9｝, which maps to eight, matching two plus six modulo ten. Check another pair before writing the general product argument.

Understand

The first isomorphism theorem states G/ker φ≅im φ. If gN=hN, then h⁻¹g∈N, so φ(g)=φ(h), proving well-definedness. The same equivalence proves injectivity in reverse. Product preservation and surjectivity onto the image complete the proof. Surjectivity onto the original codomain is not assumed.

For the second theorem, let H≤G and N normal in G. The product HN is a subgroup: normality lets an N factor pass an H factor while remaining in N, and the inverse of hn can be rewritten in HN. Restrict the quotient map to H. Its kernel is H∩N, and its image is HN/N. The first theorem therefore gives H/(H∩N)≅HN/N. This also explains why H∩N is normal in H even when H itself is not normal in G.

For the third theorem, suppose N⊆K and both are normal in G. The map G/N→G/K given by gN↦gK is well-defined because N⊆K. It is onto, with kernel K/N. Thus (G/N)/(K/N)≅G/K. Omitting the containment or normality assumptions would leave one of these quotient groups unjustified.

The subgroup correspondence is another consequence of the quotient map. Subgroups of G/N correspond to subgroups H of G containing N, by inverse image and H↦H/N. These operations are inverse: containing N ensures an entire fiber returns with any one of its elements. Inclusions and normality are preserved. This is useful for transferring a subgroup question to a smaller quotient.

Apply

For additive Z₂₀, take N=⟨10⟩ and K=⟨5⟩. Compute their sizes, then compare the two sides of the third theorem. Identify an actual map, not just two groups with the same number of elements. The quotient Z₂₀/K has five elements and is generated by the image of one.

When using a theorem, list the ambient group, subgroup, normal subgroup, map, kernel, and image in your own notation. This short audit prevents a common error: choosing the right theorem name but applying it to objects that fail its hypotheses.

Continue

Direct products next build larger groups from independent coordinates. Their element orders provide practical invariants for deciding whether a proposed isomorphism can exist.