Groups: definition, axioms, examples

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2015, 2021)
• Priority tier: T3
• Marks (count): 10 (1), 5 (1)
• Average solve time: ~5 min
• Difficulty mix: easy 2
• Section: A | Dominant type: construction / proof

Why This Chapter Matters

Group questions in UPSC Section A divide cleanly into two types: constructing composition tables for order-4 groups (2015, 5 marks) and the Bézout/well-ordering argument for the GCD (2021, 10 marks). Both are self-contained and fast: the table question rewards knowing the two groups of order 4 cold, and the Bézout question has a single 4-step proof. Once you know the axioms precisely and the well-ordering trick, both archetypes are solved in under 6 minutes.

Minimum Theory

Group axioms. A group $(G,\cdot)$ is a set $G$ with a binary operation $\cdot$ satisfying: (G1) closure: $a,b\in G\Rightarrow ab\in G$; (G2) associativity: $(ab)c=a(bc)$; (G3) identity: $\exists\, e\in G$ with $ae=ea=a$; (G4) inverses: $\forall a\in G,\,\exists\, a^{-1}$ with $aa^{-1}=a^{-1}a=e$.

Order. The order of a group $|G|$ is its cardinality. The order of an element $a\in G$ is the smallest $n\ge1$ with $a^n=e$, or $\infty$ if none exists. A group is cyclic if some element generates all of $G$: $G=\langle a\rangle=\{a^k:k\in\mathbb Z\}$.

Groups of order 4. Up to isomorphism there are exactly two: the cyclic group $\mathbb Z_4=\{e,a,a^2,a^3\}$ (has an element of order 4) and the Klein four-group $V_4=\mathbb Z_2\times\mathbb Z_2$ (every non-identity element has order 2). Any group of order 4 constructed from $\{e,a,b,c\}$ is isomorphic to one of these.

Bézout / well-ordering principle. For positive integers $m_1,\ldots,m_k$ with $\gcd = d$, the set $S=\{a_1m_1+\cdots+a_km_k : a_i\in\mathbb Z\}\cap\mathbb Z^+$ is non-empty and has a minimum $\delta$ by the well-ordering of $\mathbb Z^+$. One shows $\delta = d$, giving integers $x_1,\ldots,x_k$ with $d=x_1m_1+\cdots+x_km_k$.

Question Archetypes

construct-group-tables (1 question(s); 2015)

Recognition Cues — asked to exhibit two distinct group structures on $\{e,a,b,c\}$, one cyclic and one not; or equivalently, to write out Cayley tables for the two groups of order 4.

Solution Template

1. Identify the two groups of order 4: $\mathbb Z_4$ (cyclic) and $V_4$ (Klein, non-cyclic).
2. Write the cyclic table: powers of a generator fill each row/column.
3. Write the Klein table: every non-identity element is its own inverse; product of any two distinct non-identity elements is the third.
4. Verify both tables are Latin squares (each element appears exactly once per row and column).
5. Show cyclicity: point to an element of order 4 (cyclic case); show all non-identity orders are 2 (Klein case, hence no generator, not cyclic).

Worked Example

2015 Paper 2, 2015-P2-Q1a-ii (5 marks)

Taking a group $\{e,a,b,c\}$ of order 4, where $e$ is the identity, construct composition tables showing that one is cyclic while the other is not.

Cyclic group $\mathbb Z_4$. Relabel: $b=a^2$, $c=a^3$.

Element orders: $|e|=1$, $|a|=4$, $|b|=2$, $|c|=4$. Since $a$ has order 4, $\langle a\rangle = \{e,a,b,c\}=G$. Cyclic. $\checkmark$

Klein four-group $V_4$. Each non-identity element is its own inverse; product of any two distinct non-identity elements gives the third.

Element orders: $|e|=1$, $|a|=|b|=|c|=2$. No element generates $G$. Not cyclic. $\checkmark$

$\boxed{\text{Two groups of order 4: cyclic }\mathbb Z_4\text{ (order-4 generator) and Klein }V_4\text{ (all non-identity elements of order 2).}}$

Common Traps

• Both tables must be Latin squares: every element appears exactly once in each row and column. Check this explicitly.
• The cyclic table is not just “fill in order”; you must show the multiplication rules $a^4=e$, $a^3\cdot a=e$, etc.
• Proving non-cyclicity means showing no element of $V_4$ has order 4 — pointing to all three non-identity elements having order 2 is the correct argument.
• $\mathbb Z_4\not\cong V_4$ because the classifying invariant is the existence of an element of order 4, not the table pattern.

bezout-gcd (1 question(s); 2021)

Recognition Cues — “Show there exist integers $x_1,\ldots,x_k$ with $d = x_1m_1+\cdots+x_km_k$”; “GCD as integer linear combination”; “well-ordering principle applied to a set of linear combinations.”

Solution Template

1. Let $S = \{a_1m_1+\cdots+a_km_k : a_i\in\mathbb Z\}\cap\mathbb Z^+$. Show $S\ne\emptyset$ (take $a_1=1$, rest $0$).
2. By well-ordering, $S$ has a minimum element $\delta = x_1m_1+\cdots+x_km_k$.
3. Show $\delta\mid m_i$ for each $i$: divide $m_i = q_i\delta + r_i$; if $r_i>0$ then $r_i\in S$ with $r_i<\delta$ — contradiction. So $r_i=0$.
4. Any common divisor $d$ of all $m_i$ divides every linear combination, so $d\mid\delta$. Since $\delta$ divides each $m_i$, $\delta\le d$. Combined with $d\le\delta$, conclude $\delta=d$.

Worked Example

2021 Paper 2, 2021-P2-Q1a (10 marks)

Let $m_1,\ldots,m_k$ be positive integers, $d>0 = \gcd(m_1,\ldots,m_k)$. Show there exist integers $x_1,\ldots,x_k$ with $d = x_1m_1+\cdots+x_km_k$.

Step 1 — Set-up. Let $S = \{a_1m_1+\cdots+a_km_k : a_i\in\mathbb Z\}\cap\mathbb Z^+$. Taking $a_1=1$, $a_2=\cdots=a_k=0$ gives $m_1\in S$, so $S\ne\emptyset$.

Step 2 — Well-ordering. By the well-ordering principle, $S$ has a minimum element $\delta = x_1m_1+\cdots+x_km_k > 0$.

Step 3 — $\delta$ divides each $m_i$. Apply the division algorithm: $m_i = q_i\delta + r_i$, $0\le r_i<\delta$. Then

$r_i = m_i - q_i\delta = m_i - q_i(x_1m_1+\cdots+x_km_k),$

which is an integer linear combination of $m_1,\ldots,m_k$. If $r_i>0$ then $r_i\in S$ with $r_i<\delta$, contradicting minimality. Hence $r_i=0$, so $\delta\mid m_i$ for all $i$.

Step 4 — $\delta = d$. Since $\delta$ is a common divisor of all $m_i$, $\delta\le d$ (the gcd is the largest common divisor). Conversely, $d\mid m_i$ for all $i$, so $d$ divides every integer linear combination, in particular $d\mid\delta$, giving $d\le\delta$. Hence $\delta = d$.

$\boxed{d = x_1m_1+\cdots+x_km_k \text{ for some integers } x_i.}\;\blacksquare$

Common Traps

• Show both $\delta\le d$ and $d\le\delta$ to conclude equality — just showing $\delta$ is a common divisor is not enough.
• The key contradiction at Step 3 requires constructing the explicit expression for $r_i$ as a linear combination; write it out.
• This is the general case ($k\ge 2$ integers); for $k=2$, the standard Bézout identity $\gcd(m,n)=mx+ny$ is the special case.

Marks-Aware Writing

For the 5-mark table question: two complete, correct Cayley tables earn 3 marks; identifying orders and naming which is cyclic earns the remaining 2. Do not just draw a table without verifying the Latin square property (each element once per row/column).

For the 10-mark Bézout proof: a complete proof needs (a) non-emptiness of $S$, (b) well-ordering cited by name, (c) the division-algorithm step producing the contradiction, and (d) the two-sided argument $\delta\le d$ and $d\le\delta$. Any one of these missing costs 2–3 marks. Citing well-ordering without writing “By the well-ordering principle of $\mathbb Z^+$” is acceptable but explicit citation is safer.

Practice Set

• 2014-P2-Q1a (10 m) — — Hint: show $G$ is a group by verifying all four axioms directly, then check normality by explicit conjugation $gNg^{-1}$.