Certify a subgroup efficiently

Try

Inside the additive integers, the even integers form a smaller group. Adding or subtracting two even integers stays even, zero is even, and negating an even integer stays even. The ambient operation is unchanged. A subgroup is a subset that forms a group using that same operation, not a new operation invented for the subset.

Compare the positive even integers. They are closed under addition, but exclude zero and additive inverses. Closure alone does not certify a subgroup.

Predict

Suppose a nonempty subset H satisfies ab⁻¹∈H whenever a,b∈H. Predict how choosing a=b produces an identity, and how using that identity as a recovers inverses. The order of this argument matters because the identity has not yet been shown to belong to H.

Experiment

Consider the subgroup generated by three in additive residues modulo twelve. Repeated addition gives the following trace.

Number of additions	Residue
0	0
1	3
2	6
3	9
4	0

The generated set is ｛0,3,6,9｝. Subtract any two displayed residues and reduce modulo twelve; the result returns to the set. Check several pairs, then explain generally that differences of multiples of three are multiples of three. The general explanation supplies closure for every pair.

Understand

The one-step subgroup test says a nonempty H⊆G is a subgroup exactly when ab⁻¹∈H for all a,b∈H. Necessity follows from subgroup inverses and products. For sufficiency, choose h∈H. Then hh⁻¹=e belongs to H. Setting a=e gives b⁻¹∈H for every b∈H. Finally, for x,y∈H use b=y⁻¹ to obtain x(y⁻¹)⁻¹=xy∈H. Associativity is inherited from G. This proves every axiom rather than taking the test as an unexplained shortcut.

The intersection of any nonempty family of subgroups is a subgroup. It contains e, and membership in every subgroup is preserved by ab⁻¹. A union behaves differently. In the integers, 2Z∪3Z contains two and three but excludes their sum five. Thus unions may fail closure even when each component is a subgroup.

The subgroup generated by a subset S, written ⟨S⟩, is the intersection of all subgroups containing S. This collection is nonempty because G itself belongs to it. The intersection contains S and lies inside every subgroup containing S, making it the smallest such subgroup. Its elements can also be described as finite products of members of S and their inverses, including the empty product e.

For integer subgroups, 4Z∩6Z=12Z because an integer belongs to both exactly when it is a common multiple of four and six. The subgroup generated jointly by four and six is instead 2Z: it contains their difference two, and all integer combinations are even. Intersection and generated union answer different questions.

Apply

Compare “smallest subgroup containing both” with “largest subgroup contained in both” for 4Z and 6Z. Give a containment diagram in words or arrows, and verify each direction using divisibility rather than the sizes of the numbers alone.

Repair a proof that starts “H is closed, so H is a subgroup.” State the missing assumptions and supply either a valid subgroup-test argument or a counterexample. In a finite nonempty subset of a group, product closure does imply inverses by repeated powers, but that additional finite argument must be supplied; it is not valid for arbitrary infinite subsets.

Continue

Cyclic groups are generated by one element. Their subgroups become a divisibility problem, giving a first complete structural classification rather than a list of isolated examples.