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UPSC 2026 Maths Optional Paper 2 Q2a — Step-by-Step Solution

15 marks · Section A

Groups: definition, axioms, examples · Algebra · asked 3× in 14 yrs · Read the full method →

Question

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 $O(uv) = n$.

Technique

The recognisable trigger is two involutions: the moment a group is generated by two elements of order $2$, put $r=uv$ and $s=u$ — then $v=sr$, so $\{r,s\}$ still generates, and the pair $(r,s)$ satisfies the defining dihedral relations $r^{n}=s^{2}=e$, $srs^{-1}=r^{-1}$. From those relations alone every element collapses to the normal form $r^{k}$ or $r^{k}s$, giving $|G|\le 2n$; the whole marking weight then sits on proving those $2n$ elements are distinct, which reduces to the single claim $s\notin\langle r\rangle$. Do not merely assert the isomorphism — exhibit the bijection and check it multiplies correctly.

Solution

Step 0 — Fix the model of the dihedral group.

Write $\mathbb{Z}_n=\{0,1,\dots,n-1\}$ with addition mod $n$. Define the dihedral group of order $2n$ as the semidirect product $D_n=\mathbb{Z}_n\rtimes\mathbb{Z}_2$: as a set

with product

Put $a=(1,0)$ and $b=(0,1)$. A direct computation gives

so $D_n=\langle a,b\mid a^{n}=b^{2}=e,\ bab^{-1}=a^{-1}\rangle$ and, for $n\ge3$, $D_n$ is exactly the symmetry group of the regular $n$-gon ($n$ rotations $a^k$ and $n$ reflections $a^kb$).

Why state this first: “the dihedral group of order $2n$” must be a group we already know has order $2n$, otherwise the counting argument below is circular.

Step 1 — Choose the right generators.

We are given $u^{2}=v^{2}=e$ with $u\neq e$, $v\neq e$, and $G=\langle u,v\rangle$ finite. Set

Since $G$ is finite, $r$ has a finite order; by hypothesis

Step 2 — $\{r,s\}$ generates $G$.

Clearly $r,s\in G$, so $\langle r,s\rangle\subseteq G$. Conversely $u=s$ and, since $u^{2}=e$,

So both $u$ and $v$ lie in $\langle r,s\rangle$, whence $G=\langle u,v\rangle\subseteq\langle r,s\rangle$. Therefore

Step 3 — Verify the three dihedral relations.

1. $r^{n}=e$ by the definition of $n=O(r)$, and no smaller positive power of $r$ is $e$.
2. $s^{2}=u^{2}=e$, and $s=u\neq e$, so $O(s)=2$.
3. Using $u^{-1}=u$ and $v^{-1}=v$,

Hence

Iterating $(\ast)$: $s r^{k}s^{-1}=(srs^{-1})^{k}=r^{-k}$, and since $s^{-1}=s$,

Relation $(\ast\ast)$ is the working form: it lets every $s$ be pushed to the right past any power of $r$.

Step 4 — Every element of $G$ has the normal form $r^{k}$ or $r^{k}s$.

Let

We show $H$ is a subgroup. All indices are read mod $n$, legitimate because $r^{n}=e$. Using $(\ast\ast)$ and $s^{2}=e$:

So $H$ is closed under multiplication. Also $e=r^{0}\in H$; and $H$ is closed under inverses, since $(r^{k})^{-1}=r^{n-k}\in H$ and

(so incidentally every element $r^{k}s$ is an involution — the “reflections”).

Thus $H\le G$. Moreover $u=s=r^{0}s\in H$, and by Step 2 together with $(\ast\ast)$, $v=sr=r^{-1}s\in H$. Since $G=\langle u,v\rangle$ is the smallest subgroup containing $u$ and $v$, we get $G\subseteq H$; and $H\subseteq G$ is clear. Hence

Step 5 — The key lemma: $s\notin\langle r\rangle$.

This is where the marks are. Suppose, for contradiction, that $s\in\langle r\rangle$. Then $G=\langle r,s\rangle=\langle r\rangle$ is cyclic, hence abelian. Being abelian, $u$ and $v$ commute, so

and therefore $n=O(r)\in\{1,2\}$.

• If $n=1$: $\langle r\rangle=\{e\}$, so $s=e$, i.e. $u=e$ — contradicting $O(u)=2$.
• If $n=2$: $\langle r\rangle=\{e,r\}$ and $s\neq e$, so $s=r$, i.e. $u=uv$, giving $v=e$ — contradicting $O(v)=2$.

Both cases are impossible. Hence

Step 6 — The $2n$ listed elements are distinct, so $|G|=2n$.

Three checks, and all three are needed:

1. Rotations are distinct. If $r^{i}=r^{j}$ with $0\le i<j\le n-1$ then $r^{\,j-i}=e$ with $0<j-i<n$, contradicting $O(r)=n$.
2. Reflections are distinct. If $r^{i}s=r^{j}s$ then right-multiplying by $s^{-1}$ gives $r^{i}=r^{j}$, so $i=j$ by (1).
3. No rotation equals a reflection. If $r^{i}=r^{j}s$ for some $i,j$, then $s=r^{\,i-j}\in\langle r\rangle$, contradicting (5.1).

So the list in (4.2) has exactly $2n$ distinct entries and

(Structural aside, worth one line: $\langle r\rangle$ has index $2$ in $G$ by (6.1), hence $\langle r\rangle\trianglelefteq G$, and $(\ast)$ says $s$ acts on it by inversion — i.e. $G=\langle r\rangle\rtimes\langle s\rangle$, which is precisely the shape of $D_n$ in Step 0.)

Step 7 — Construct the isomorphism explicitly.

By Step 0 the $2n$ elements $a^{k},a^{k}b$ $(0\le k\le n-1)$ of $D_n$ are distinct and exhaust $D_n$; by Step 6 the $2n$ elements $r^{k},r^{k}s$ are distinct and exhaust $G$. So the assignment

is well defined (each element of $G$ has exactly one normal form) and is a bijection (it matches the two lists term by term).

$\varphi$ is a homomorphism: in $D_n$ the relations $a^{n}=b^{2}=e$, $b a^{j}=a^{-j}b$ give, exactly as in (4.1),

Comparing with (4.1) case by case:

These four cases cover every product in $G$. Hence $\varphi$ is a bijective homomorphism, i.e.

(Alternative one-line finish, if von Dyck’s theorem is available: $r,s\in G$ satisfy the defining relations of $D_n=\langle a,b\mid a^{n}=b^{2}=e,\,bab^{-1}=a^{-1}\rangle$, so there is a homomorphism $D_n\to G$ with $a\mapsto r$, $b\mapsto s$; it is onto because $\langle r,s\rangle=G$ (Step 2), and $|D_n|=2n=|G|$ by Step 6, so it is injective as well. Steps 5–6 are still needed — they supply $|G|=2n$ — so nothing is saved except the four-case check.)

Step 8 — The small cases (honesty at the boundary).

The proof above is uniform in $n\ge1$, but it is worth recording what the conclusion means at the two degenerate values, since “dihedral” is usually pictured as a polygon group:

• $n=1$. $uv=e\Rightarrow v=u^{-1}=u$. The two generators coincide, $G=\langle u\rangle\cong\mathbb{Z}_2$, of order $2n=2$. Consistently, $D_1=\mathbb{Z}_1\rtimes\mathbb{Z}_2\cong\mathbb{Z}_2$.
• $n=2$. $(uv)^{2}=e$ gives $uvuv=e$, so $uv=(uv)^{-1}=vu$: $u$ and $v$ commute, and $u\neq v$. Then $G=\{e,u,v,uv\}$ is the Klein four-group $\mathbb{Z}_2\times\mathbb{Z}_2$, of order $2n=4$ — and indeed $D_2\cong\mathbb{Z}_2\times\mathbb{Z}_2$. Note this is abelian: “dihedral of order 4” is the Klein group, not a polygon symmetry group.
• $n\ge3$. $G$ is non-abelian (from $(\ast)$, $sr=r^{-1}s\neq rs$ since $r^{2}\neq e$) and is the genuine symmetry group of the regular $n$-gon.

In particular $G$ is abelian $\iff n\le2$.

Step 9 — Where finiteness was used.

Finiteness of $G$ was used exactly once — to guarantee $n=O(uv)<\infty$ in Step 1. It is not an extra assumption beyond that: by (6.1), $|G|=2\,O(uv)$, so $G$ is finite iff $uv$ has finite order. If $O(uv)=\infty$ the same Steps 2–4 give $G\cong D_\infty=\langle a,b\mid b^{2}=e,\ bab^{-1}=a^{-1}\rangle$, the infinite dihedral group.

Answer