UPSC Maths 2025 Paper 2 Q1a — Solution

10 marks · Section A

Question

Let $H$ and $K$ be two subgroups of a group $G$ such that $o(H) > \sqrt{o(G)}$ and $o(K) > \sqrt{o(G)}$ . Show that $H \cap K \neq \{e\}$ , where $e$ is the identity element. Here $o(H)$ , $o(K)$ and $o(G)$ denote the order of $H$ , $K$ and $G$ respectively.

Technique

Count using the product-set order formula $|HK| = \dfrac{|H|\,|K|}{|H\cap K|}$ together with the bound $|HK| \le |G|$ .

Solution

Since $H,K$ are finite subgroups of $G$ , consider the product set $HK = \{\, hk : h\in H,\ k\in K \,\}.$

Step 1 — Order of the product set. For finite subgroups $H,K$ of $G$ , $|HK| = \frac{|H|\,|K|}{|H\cap K|}. \tag{$\ast$}$

Proof of $(\ast)$ . Define a map from $H\times K$ onto $HK$ by $(h,k)\mapsto hk$ . For a fixed product $g=hk$ , the pairs $(h',k')$ with $h'k'=hk$ are exactly $h' = ht,\ k' = t^{-1}k$ for $t\in H\cap K$ (since $h'k'=hk \Rightarrow h^{-1}h' = k(k')^{-1} =: t \in H\cap K$ ). Thus every element of $HK$ has exactly $|H\cap K|$ preimages, so $|H\times K| = |HK|\cdot |H\cap K| \implies |HK| = \frac{|H|\,|K|}{|H\cap K|}.$

Step 2 — Bound the product set. Although $HK$ need not be a subgroup (since $G$ may be non-Abelian), it is a subset of $G$ , so $|HK| \le |G| = o(G). \tag{$\ast\ast$}$

Step 3 — Combine. Suppose, for contradiction, that $H\cap K = \{e\}$ , i.e. $|H\cap K| = 1$ . Then by $(\ast)$ , $|HK| = |H|\,|K| = o(H)\,o(K).$ Using the hypotheses $o(H) > \sqrt{o(G)}$ and $o(K) > \sqrt{o(G)}$ , $|HK| = o(H)\,o(K) > \sqrt{o(G)}\cdot\sqrt{o(G)} = o(G).$ This contradicts $(\ast\ast)$ , namely $|HK| \le o(G)$ .

Hence the assumption is false, and $|H\cap K| > 1, \quad\text{i.e.}\quad H\cap K \neq \{e\}. \qquad \blacksquare$

Answer

$H\cap K \neq \{e\}$ . The proof rests on $|HK| = \dfrac{o(H)\,o(K)}{|H\cap K|} \le o(G)$ ; if the intersection were trivial, $o(H)\,o(K) > o(G)$ would force $|HK| > o(G)$ , which is impossible.