Subgroups and the Subgroup Criterion

Subgroups and the tests for recognizing them are the entry point to nearly all subsequent structural theory in algebra. Cosets and Lagrange's theorem (foreshadowed here) use subgroups to partition a finite group into equal-size blocks, giving the fundamental divisibility constraint $|H|$ divides $|G|$. Kernels of group homomorphisms — central to ring theory, module theory, and Galois theory — are always subgroups, verified by exactly the one-step criterion applied to preimages of the identity. In cryptography, the security of Diffie–Hellman key exchange and elliptic-curve cryptosystems rests on the subgroup structure of multiplicative groups modulo a prime or of points on an elliptic curve, where cyclic subgroups of specific orders are deliberately chosen. In physics, subgroups of symmetry groups (e.g. discrete rotation subgroups of $O(3)$ describing crystal symmetry, or the subgroups of the Lorentz group corresponding to physically realizable transformations) determine selection rules and conserved quantities. In computer science, subgroup membership testing underlies algorithms for permutation groups (e.g. Schreier–Sims methods) used in computational group theory software.

Ask the tutor: Can you show me why the intersection of two subgroups is always a subgroup, using the two-step test?

Think of a group $G$ as a symmetry pattern — say, the rotations and reflections of a regular hexagon, forming the dihedral group of order 12. A subgroup is a self-contained sub-pattern: a subset of symmetries that, whenever you apply two of them in a row, or undo one, you land back inside that same subset. The hexagon's rotational symmetries alone (ignoring reflections) form such a sub-pattern, isomorphic to the cyclic group of order 6. So does the pair {identity, one specific 180° rotation}. So does the trivial group consisting of the identity alone. Not every subset works, though: the set of just the six reflections is not closed, because composing two reflections produces a rotation, not a reflection. The subtlety the subgroup criterion resolves is this: verifying "is a subset $H$ a group in its own right?" naively requires checking four separate axioms. But three of those checks are either automatic or logically entangled with each other. Associativity holds for all triples in $G$, hence certainly for triples drawn from $H \subseteq G$ — it costs nothing to re-verify. And it turns out that if $H$ is nonempty and closed under forming $ab^{-1}$ from any two of its elements, the identity and all inverses fall out as forced consequences. This is why the one-step test packages everything into a single implication about the map $(a,b) \mapsto ab^{-1}$.

Example 1 (one-step test on a genuinely asymmetric subset). Let $G = (\mathbb{Z}/24\mathbb{Z}, +)$, and let

We verify $H \le G$ using the one-step test (written additively, so the test becomes: $H \ne \varnothing$, and $a,b \in H \Rightarrow a - b \in H$). Nonemptiness. $0 \in H$ since $3 \cdot 0 = 0 \equiv 0 \pmod{24}$. So $H \ne \varnothing$. Closure under $a - b$. We check all ordered pairs from $\{0,8,16\}$ directly, since $H$ is small and explicit:

• $0 - 0 = 0 \in H$
• $0 - 8 = -8 \equiv 16 \pmod{24} \in H$
• $0 - 16 = -16 \equiv 8 \pmod{24} \in H$
• $8 - 0 = 8 \in H$
• $8 - 8 = 0 \in H$
• $8 - 16 = -8 \equiv 16 \pmod{24} \in H$
• $16 - 0 = 16 \in H$
• $16 - 8 = 8 \in H$
• $16 - 16 = 0 \in H$

Every difference lies in $\{0,8,16\} = H$. By the one-step test, $H \le \mathbb{Z}/24\mathbb{Z}$. Cross-check via structure. Indeed $H = 8\mathbb{Z}/24\mathbb{Z}$, the multiples of $8$ modulo $24$, which is exactly the unique subgroup of order $3$ inside the cyclic group of order $24$ — consistent with the fact that cyclic groups have exactly one subgroup of each order dividing the group order (a fact developed further in Cyclic Groups and Generators). This independently confirms our direct computation.

Example 2 (finite subset test inside a nonabelian group, avoiding a fake shortcut). Let $G = GL_2(\mathbb{R})$, the group of invertible $2\times 2$ real matrices under matrix multiplication, and consider

Here $A$ is rotation by $90°$. We first compute the powers explicitly to know $H$ concretely, then apply the finite subset test (closure alone suffices, since $H$ is finite). Compute:

So $H = \{I, A, -I, A^3\}$ has exactly four distinct elements (one checks directly that $I, A, -I, A^3$ are pairwise distinct matrices, e.g. their $(1,1)$-entries are $1, 0, -1, 0$ respectively and the $(1,2)$-entries of the two entries with $(1,1)=0$ are $-1$ and $1$, distinguishing $A$ from $A^3$). Closure. Products of elements of $H$ are again powers of $A$ with exponent taken mod $4$: for $A^i, A^j \in H$ with $0 \le i,j \le 3$, $A^i A^j = A^{i+j}$, and since $A^4 = I$, this equals $A^{(i+j)\bmod 4} \in H$. We verify a representative nontrivial case explicitly: $A^3 \cdot A^3 = A^6 = A^4 \cdot A^2 = I \cdot A^2 = A^2 = -I \in H$. So $H$ is closed under multiplication. Nonemptiness and finiteness. $H$ is nonempty (it contains $I$) and has exactly $4$ elements, hence is finite. By the Finite Subset Test, $H \le GL_2(\mathbb{R})$ — with no need to separately verify inverses. (As a sanity check, one can see directly that $A^{-1} = A^3$ since $A\cdot A^3 = A^4 = I$, and $A^3 \in H$, consistent with the theorem's guarantee.) This subgroup $H$ is cyclic of order $4$, generated by the rotation matrix $A$, and it is a subgroup of the infinite, nonabelian group $GL_2(\mathbb{R})$ — illustrating that the finite subset test applies even when the ambient group is infinite and noncommutative, as long as the subset itself is finite.