Groups: definition, axioms, examples

At a Glance

• Frequency: 3 sub-parts across 3 of 14 years (2015, 2021, 2026)
• 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.

The dihedral group $D_n$, and presentations. $D_n$ is the group of order $2n$ with

$D_n=\langle r,s \mid r^n=s^2=e,\ srs^{-1}=r^{-1}\rangle=\{r^k,\,r^ks : 0\le k\le n-1\}.$

Concretely it is $\mathbb{Z}_n\rtimes\mathbb{Z}_2$, with product $(k,\varepsilon)(l,\delta)=(k+(-1)^\varepsilon l,\ \varepsilon+\delta)$; for $n\ge3$ it is the symmetry group of the regular $n$-gon ($n$ rotations $r^k$, $n$ reflections $r^ks$). Two conventions that must be stated when they are used: $D_1\cong\mathbb{Z}_2$ and $D_2\cong\mathbb{Z}_2\times\mathbb{Z}_2$ (abelian — “dihedral of order 4” is the Klein group, not a polygon group). $G$ is abelian $\iff n\le2$.

The working form of the relation is obtained by iterating $srs^{-1}=r^{-1}$ and using $s^{-1}=s$:

$s\,r^k=r^{-k}\,s\qquad\text{for every }k\in\mathbb{Z},$

which lets every $s$ be pushed to the right past any power of $r$ — that is what produces the normal form $r^k$ or $r^ks$ and hence the bound $|G|\le2n$.

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.

dihedral-from-involutions (1 question(s); 2026)

Two involutions is the trigger. The instant a group is generated by two elements of order 2, set $r=uv$ and $s=u$: the pair $(r,s)$ satisfies the defining dihedral relations, and the question becomes a counting problem.

Recognition Cues

• ”$G$ is generated by two elements $u,v$ of order 2 each.”
• The conclusion names $D_n$ with $n=O(uv)$ — the order of the product, not of either generator.
• “Finite group” is given, and is doing exactly one job (see the traps).

Solution Template

1. Fix a model of $D_n$ first, one already known to have order $2n$ — otherwise the counting argument at the end is circular.
2. Change generators: $r=uv$, $s=u$. Then $v=sr$ (since $u^2=e$), so $\langle r,s\rangle=\langle u,v\rangle=G$.
3. Verify the three relations: $r^n=e$ by definition of $n=O(r)$; $s^2=u^2=e$; and, using $u^{-1}=u$, $v^{-1}=v$, $srs^{-1}=u(uv)u=vu=(uv)^{-1}=r^{-1}.$
4. Normal form: show $H=\{r^k\}\cup\{r^ks\}$ is closed under products and inverses using $sr^k=r^{-k}s$, contains $u$ and $v$, hence equals $G$. This gives $|G|\le 2n$.
5. Prove $s\notin\langle r\rangle$ — this is where the marks are. If $s\in\langle r\rangle$ then $G=\langle r\rangle$ is cyclic, hence abelian, so $u,v$ commute and $r^2=(uv)^2=u^2v^2=e$, forcing $n\le2$; both $n=1$ and $n=2$ then contradict $O(u)=O(v)=2$.
6. Distinctness, three checks: rotations distinct (order of $r$); reflections distinct (cancel $s$); no rotation equals a reflection (else $s\in\langle r\rangle$). Hence $|G|=2n$.
7. Exhibit the isomorphism $\varphi(r^k)=a^k$, $\varphi(r^ks)=a^kb$ and check it respects the four product types — do not merely assert it.

Worked Example

2026 Paper 2, 2026-P2-Q2a (15 marks)

Let $G$ be a finite group generated by two elements $u$ and $v$ of order 2 each. Show that $G$ is isomorphic to the dihedral group of order $2n$, where $O(uv)=n$.

Source: analysis/solutions/2026-P2-Q2a.md

Generators. Put $r=uv$, $s=u$, $n=O(r)$ (finite because $G$ is). Then $sr=u(uv)=u^2v=v$, so $u,v\in\langle r,s\rangle$ and $G=\langle r,s\rangle$.

Relations. $r^n=e$; $s^2=u^2=e$ with $s\ne e$; and $srs^{-1}=u(uv)u^{-1}=vu=(uv)^{-1}=r^{-1}$. Iterating, $sr^k=r^{-k}s$.

Normal form. With $H=\{r^k\}_{k=0}^{n-1}\cup\{r^ks\}_{k=0}^{n-1}$, the four products are

$r^ir^j=r^{i+j},\quad r^i(r^js)=r^{i+j}s,\quad (r^is)r^j=r^{i-j}s,\quad (r^is)(r^js)=r^{i-j},$

so $H$ is closed; $(r^ks)^2=e$ gives closure under inverses. Since $u=s\in H$ and $v=r^{-1}s\in H$ and $G=\langle u,v\rangle$ is the smallest such subgroup, $G=H$ and $|G|\le2n$.

The key lemma. If $s\in\langle r\rangle$ then $G=\langle r\rangle$ is cyclic, hence abelian, so $r^2=(uv)(uv)=u(vu)v=u(uv)v=u^2v^2=e$ and $n\in\{1,2\}$. If $n=1$ then $s=e$, i.e. $u=e$ — contradicting $O(u)=2$. If $n=2$ then $\langle r\rangle=\{e,r\}$ and $s\ne e$ force $s=r$, i.e. $u=uv$, giving $v=e$ — contradicting $O(v)=2$. Hence $s\notin\langle r\rangle$.

Distinctness. $r^i=r^j$ with $0\le i<j<n$ would give $r^{j-i}=e$, impossible; $r^is=r^js$ cancels to $r^i=r^j$; and $r^i=r^js$ would put $s=r^{i-j}\in\langle r\rangle$. So the $2n$ listed elements are distinct and $|G|=2n$.

Isomorphism. $\varphi(r^k)=a^k$, $\varphi(r^ks)=a^kb$ is well defined and bijective (it matches two lists of $2n$ distinct elements term by term), and the four product identities above match the corresponding ones in $D_n$ case by case. Hence $G\cong D_n$. $\blacksquare$

The boundary cases, worth 90 seconds. $n=1$: $uv=e$ so $v=u$ and $G=\langle u\rangle\cong\mathbb{Z}_2=D_1$. $n=2$: $u,v$ commute and $G=\{e,u,v,uv\}$ is the Klein four-group $=D_2$, which is abelian. For $n\ge3$, $G$ is non-abelian and is the symmetry group of the regular $n$-gon.

Common Traps

• Asserting $|G|=2n$. The normal form only gives $|G|\le2n$; the reverse needs $s\notin\langle r\rangle$ plus the three distinctness checks. Skipping this drops 4–5 marks.
• Writing ”$G$ is non-abelian, being dihedral”. False at $n=1,2$. The statement is true as printed for every $n\ge1$ only under the conventions $D_1\cong\mathbb{Z}_2$, $D_2\cong\mathbb{Z}_2\times\mathbb{Z}_2$.
• Assuming $u\ne v$. The question does not say it; $u=v$ gives $n=1$ and must not break the proof.
• Exhibiting a bijection and stopping. An isomorphism must be checked to respect the operation — the four product types are the check.
• Misreading which order is $n$. It is $O(uv)$, not $O(u)$ or $O(v)$ (both are 2). Harmlessly, $O(uv)=O(vu)$, since $vu=u(uv)u^{-1}$ is conjugate to $uv$.
• “Finite” is not decorative but it is small. It is used exactly once, to make $O(uv)$ finite. Since $|G|=2\,O(uv)$, the hypothesis is equivalent to ”$uv$ has finite order”; dropping it gives the infinite dihedral group $D_\infty$, not a counterexample.

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 $gNg^{-1}$.