Subgroup Definition and Verification Techniques: Worked Examples for GATE CS

A subset may contain the identity and still fail to be a subgroup. The definition test, the one-step test, and a complete operation table may look like three separate methods, but each checks the same group requirements. The subgroup H={0,3,6,9} inside (Z_12,+) can be verified by all three routes, and counterexamples show where shortcuts fail. If closure notation is unfamiliar, review Set Theory and Relations for GATE: Closures and Posets first.

Subgroup definition: the set and operation must stay together

Writing H <= G means that G is a group under a stated operation *, H is a non-empty subset of G, and H is itself a group under that same restricted operation. Therefore, H must have closure, associativity, an identity, and inverses. Associativity is inherited from G, so direct verification checks non-emptiness, closure, the identity, and inverses.

Containing the identity is not enough. In (Z_8,+), A={0,2,4} contains 0, but 2+4=6 mod 8, and 6 is not in A. In contrast, B={0,4} is a subgroup: its four sums are 0+0=0, 0+4=4, 4+0=4, and 4+4=0 modulo 8. Both elements are self-inverse.

For additive groups, the identity is 0 and the inverse of a is -a. For multiplicative groups, the identity is e or 1, and the inverse is a^-1. The wider map is covered in Group Theory: Groups, Rings and Fields for GATE CS.

Subgroup verification techniques: choose the shortest valid proof

Each technique below is a complete proof only when all its stated conditions hold.

The one-step test really includes the missing checks. For a in H, choosing a,a gives aa^-1=e in H. Choosing e,a then gives ea^-1=a^-1 in H. Since b^-1 is in H, applying the condition to a,b^-1 gives a(b^-1)^-1=ab in H.

For 4Z <= Z under addition, take arbitrary 4m,4n in 4Z. Then (4m)-(4n)=4(m-n), and m-n is an integer, so the difference stays in 4Z. Checking several numerical pairs would only suggest this infinite pattern, not prove it.

Worked subgroup example: verify {0,3,6,9} inside Z_12

First fix the objects. Let G=Z_12={0,1,...,11} under addition modulo 12, and let H={0,3,6,9}. The set is non-empty, and it contains the identity 0.

Closure can be checked completely because H is finite. Every entry below is the row element plus the column element, reduced modulo 12.

Every result belongs to H, so closure holds. The cells whose sum is 0 give the inverses: -0=0, -3=9, -6=6, and -9=3 modulo 12. Associativity is inherited from addition modulo 12; inspecting selected table cells would not prove it. All subgroup conditions are now accounted for, so H <= Z_12.

The additive subgroup test compresses the proof. Write H={3k mod 12 : k in Z}. If a=3r mod 12 and b=3s mod 12, then a-b=3(r-s) mod 12, which is again in H. Since H is non-empty, the test proves the result.

There are also two useful structural checks. Repeated addition of 3 gives <3>={0,3,6,9}, so H is cyclic of order 4. Its distinct cosets are H={0,3,6,9}, 1+H={1,4,7,10}, and 2+H={2,5,8,11}. They are disjoint and together contain all 12 residues.

Subgroup examples and non-examples across common operations

The infinite additive example 4Z <= Z follows from 4m-4n=4(m-n). For a finite multiplicative example, consider U(8)={1,3,5,7} under multiplication modulo 8. For K={1,7}, the product table has rows 1,7 and 7,1: 1x1=1, 1x7=7, 7x1=7, and 7x7=49, which is congruent to 1 mod 8. Thus 1 is the identity, and both elements have inverses in K.

A single witness can reject a candidate. The set A={0,2,4} in (Z_8,+) fails closure because 2+4=6 mod 8. The positive integers P={1,2,3,...} are not a subgroup of the positive rationals under multiplication because the inverse of 2 is 1/2, which is not in P.

The same-looking carrier can behave differently under another operation. A correct solution always names the ambient group, the operation, and the satisfied or failed axiom. It also confirms that the proposed ambient set is itself a group.

Subgroup traps: necessary conditions are not complete proofs

The claim "H contains the identity, so it is a subgroup" is false. In Z_3, {0,1} contains 0, but 1+1=2 lies outside the subset. The claim "|H| divides |G|, so H is a subgroup" reverses Lagrange's theorem. The set {0,1} has size 2, which divides |Z_4|=4, yet 1+1=2 lies outside it.

For finite groups, Lagrange's theorem is a rejection screen: if H <= G, then |H| divides |G|. A five-element candidate cannot be a subgroup of a group of order 12. A four-element candidate passes that screen but still needs verification.

Always use the stated operation. Ordinary addition is irrelevant if the operation is addition modulo n. A Cayley table can display closure, identity, and inverse candidates, but associativity must come from the ambient group or a separate proof.

Unions create another trap. In Z_12, H=<3>={0,3,6,9} and K=<4>={0,4,8} are subgroups. Their union is not: 3+4=7 belongs to neither set and hence is outside the union.

How exam-style subgroup questions test the same checks

Common question forms include direct verification, finding a counterexample, determining a generated subgroup, applying an order screen, and deciding whether intersections or unions remain subgroups. The GATE CS Exam Preparation category connects this topic to the wider discrete-mathematics syllabus.

Four rapid checks capture the pattern:

1. In Z_12, <4>={0,4,8}, so its order is 3.

2. A subgroup of a group of order 12 cannot have order 5, by Lagrange's theorem.

3. For H=<3> and K=<4> in Z_12, H intersection K={0}, which is a subgroup.

4. Their union fails closure because 3+4=7 lies outside it.

Under time pressure, first write the operation and identity. Next try the one-step test, and stop as soon as a counterexample appears. Use Lagrange's theorem only as a filter. Then state the conclusion with the exact condition proved or violated. Practise the same decisions in Discrete Mathematics MCQs, which mixes subgroup checks with other discrete-mathematics questions.

Subgroup verification checklist and the next step

The short version is to confirm the ambient group and operation, show the candidate is non-empty, then use the definition test or prove ab^-1 in H. Reject with one valid counterexample, and treat order divisibility only as a necessary screen. Here, {0,3,6,9} <= Z_12 because it contains 0, stays closed modulo 12, has inverses 0,9,6,3, and inherits associativity. For a structured route through Discrete Mathematics and Group Theory, consider GATE Guidance by Sanchit Sir.