Count groups by equal-sized pieces

Try

In additive residues modulo twelve, take H=｛0,3,6,9｝. Add one to every member to form 1+H, then add two to form 2+H. These translated sets cover all twelve residues without overlap. Adding three produces H again, so a new representative need not produce a new set.

For a general group, a left coset is aH=｛ah:h∈H｝. It is a subset of G, not usually a subgroup. Only the identity coset H must contain the identity element.

Predict

Predict whether two left cosets can overlap in just one element without being equal. Try to build such an overlap in the residue example. The failed search suggests a theorem, but the proof needs arbitrary group elements and must not rely on commutativity.

Experiment

Organize the residue example as three rows. Each row has the same number of entries because addition by a fixed residue is reversible.

Representative	Coset
0	｛0,3,6,9｝
1	｛1,4,7,10｝
2	｛2,5,8,11｝

Check that 4+H equals 1+H. Their representative difference is three, an element of H. In multiplicative notation, the corresponding criterion is aH=bH exactly when b⁻¹a∈H. The side of the inverse is important in a nonabelian group.

Understand

If aH=bH, then a belongs to bH, so a=bh for some h∈H and b⁻¹a∈H. Conversely, if a=bh with h∈H, then aH=bhH=bH because hH=H. This proves the equality criterion in both directions.

If two cosets intersect, write ah₁=bh₂ for an element of the intersection. Then b⁻¹a=h₂h₁⁻¹∈H, so the cosets are equal. Every g lies in gH, since e∈H. Therefore left cosets partition G. The map h↦ah is a bijection from H to aH, with inverse x↦a⁻¹x, so every coset has the same size.

For finite G, counting this partition gives Lagrange’s theorem: |G|=[G:H]|H|, where the index [G:H] is the number of left cosets. In particular, |H| divides |G|. Applying this to the cyclic subgroup generated by g proves that an element’s order divides the group order. A group of prime order is consequently cyclic: any nonidentity element has order greater than one dividing that prime.

The converse of Lagrange is false in general. Divisibility does not automatically construct a subgroup for each divisor. For example A₄ has no subgroup of order six, though six divides twelve. A proof of that claim will become available from normality and conjugacy classes: an index-two subgroup would be normal, but the class sizes cannot form such a subgroup. Do not use an unproved converse while waiting for those tools.

Apply

Given a group of order thirty, list the element orders that Lagrange permits. Label them “possible by divisibility,” not “known to occur.” Then choose a particular cyclic group of that order and explain why its subgroup theorem gives more information.

Contrast left and right cosets in a noncommutative example. Each forms a partition with equal sizes, but the sets aH and Ha may differ. Counting them does not justify multiplying cosets as if they were elements of a quotient group.

Continue

Normality next identifies exactly when the familiar coset multiplication is independent of representatives. The partition theorem supplies the underlying set; an additional compatibility condition supplies the group operation.