Subgroups; Subgroup Criterion

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2021)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~18 min
• Difficulty mix: medium 1
• Section: B | Dominant type: proof

Why This Chapter Matters

Subgroups are the fundamental building blocks of group theory, and the subgroup criterion appears in virtually every abstract algebra course as the first non-trivial theorem. UPSC 2021 tested the ability to verify that a given subset is a subgroup using the one-step or two-step criterion — a 15-mark Section B question demands a rigorous proof with every detail justified. Mastering this criterion also underpins later results such as Lagrange’s theorem, normal subgroups, and quotient groups.

Minimum Theory

Definition

Let $(G, \cdot)$ be a group. A non-empty subset $H \subseteq G$ is a subgroup of $G$, written $H \le G$, if $H$ is itself a group under the operation inherited from $G$.

Three-Step Criterion

$H \le G$ if and only if:

1. $H \ne \emptyset$,
2. $a, b \in H \Rightarrow ab \in H$ (closure under the operation),
3. $a \in H \Rightarrow a^{-1} \in H$ (closure under inverses).

Proof that these suffice. Since $H \ne \emptyset$, pick $a \in H$. By (3), $a^{-1} \in H$. By (2), $a \cdot a^{-1} = e \in H$, so the identity is present. Associativity is inherited from $G$. Hence $H$ is a group.

One-Step Criterion (most efficient for proofs)

$H \le G \iff H \ne \emptyset \text{ and } \forall\, a, b \in H,\; ab^{-1} \in H.$

Proof. ($\Rightarrow$) Clear. ($\Leftarrow$) Take $a = b$: $aa^{-1} = e \in H$. Take $a = e$: $eb^{-1} = b^{-1} \in H$. Then for $a, b \in H$ we have $b^{-1} \in H$, so $a(b^{-1})^{-1} = ab \in H$.

Finite Subgroup Criterion

If $G$ is a finite group then $H \le G$ if and only if $H \ne \emptyset$ and $H$ is closed under the operation. (Inverses come for free by finiteness: the powers $a, a^2, a^3, \ldots$ eventually return to $e$, giving $a^{-1}$ as a power of $a$.)

Standard Examples

Center of $G$. $Z(G) = \{ g \in G : gx = xg \text{ for all } x \in G \}.$ $Z(G) \le G$: it is non-empty ($e \in Z(G)$); if $a, b \in Z(G)$ then $(ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab)$, so $ab \in Z(G)$; if $a \in Z(G)$ then $ax = xa \Rightarrow a^{-1}x = xa^{-1}$ (multiply both sides on left and right by $a^{-1}$), so $a^{-1} \in Z(G)$.

Cyclic subgroup generated by $a$. $\langle a \rangle = \{ a^n : n \in \mathbb{Z} \}.$ This is always a subgroup: $a^m \cdot a^n = a^{m+n} \in \langle a \rangle$ and $(a^n)^{-1} = a^{-n} \in \langle a \rangle$.

Intersection of subgroups. If $H, K \le G$ then $H \cap K \le G$. (The union $H \cup K$ need not be a subgroup.)

Question Archetypes

verify-subgroup (1 question; 2021)

Recognition Cues

• Prompt says “show that $H$ is a subgroup of $G$” or “verify that $H \le G$”
• A specific set is defined — often with an algebraic condition
• 15 marks (Section B): full rigorous proof expected

Solution Template

1. Non-emptiness. Exhibit a concrete element (usually $e$ or $0$).
2. Closure under inverses. For arbitrary $a \in H$, show $a^{-1} \in H$.
3. Closure under operation. For arbitrary $a, b \in H$, show $ab \in H$.

(Alternatively, use the one-step criterion: show $ab^{-1} \in H$ directly.)

4. Conclude: $H \le G$ by the subgroup criterion. $\blacksquare$

Worked Example

2021 Paper 2, 2021-P2-Q2b (15 marks)

Let $G = (\mathbb{R} \setminus \{0\}, \cdot)$ be the multiplicative group of non-zero real numbers. Let $H = \{ x \in \mathbb{R} : x > 0 \}.$ Show that $H$ is a subgroup of $G$.

Solution.

We apply the one-step subgroup criterion.

Step 1: $H \ne \emptyset$. We have $1 > 0$, so $1 \in H$.

Step 2: Closed under $ab^{-1}$. Let $a, b \in H$, i.e., $a > 0$ and $b > 0$.

In $G$ the operation is multiplication, so $b^{-1} = 1/b$. Since $b > 0$ we have $1/b > 0$, and then $ab^{-1} = \frac{a}{b} > 0,$ so $ab^{-1} \in H$.

Conclusion. By the one-step criterion, $H \le G$.

$\boxed{H \le G}\;\blacksquare$

Common Traps

• Forgetting to verify non-emptiness — always exhibit $e \in H$ explicitly.
• Applying the finite subgroup criterion to an infinite group.
• Confusing $H \cup K \le G$ (false in general) with $H \cap K \le G$ (always true).
• For the one-step criterion, writing $a^{-1}b$ instead of $ab^{-1}$ — the order matters in non-abelian groups.

Marks-Aware Writing

15 marks (Section B): The examiner expects:

• A clearly stated version of whichever criterion you use (1–2 marks).
• Non-emptiness shown by example (1–2 marks).
• Closure under inverses with calculation (4–5 marks).
• Closure under the operation with calculation (4–5 marks).
• Explicit concluding sentence citing the criterion (1 mark).

Do not just say “clearly closed” — compute and justify each step.

Practice Set

Only one historical question on this atom (shown above).