UPSC 2023 Maths Optional Paper 2 Q1a — Step-by-Step Solution

10 marks · Section A

Group homomorphisms: kernel, image · Algebra · asked 4× in 14 yrs · Read the full method →

Question

Let $G$ be a group of order 10 and $G'$ be a group of order 6. Examine whether there exists a homomorphism of $G$ onto $G'$ .

First Isomorphism Theorem + Lagrange (image order divides domain order).

Strategy. Apply the First Isomorphism Theorem; the order of the image must divide $|G|$ .

Step 1 — First Isomorphism Theorem.

Suppose $\phi:G\to G'$ is a surjective homomorphism. Let $K=\ker\phi\trianglelefteq G$ . Then

G/K\;\cong\;\phi(G)\;=\;G'.

Step 2 — Order constraint.

By Lagrange’s theorem $|G/K|=|G|/|K|$ , so $|G/K|$ divides $|G|=10$ . Combined with Step 1,

|G'|=|G/K|\;\big|\;|G|,\quad\text{i.e.,}\quad 6\mid 10.

Step 3 — Contradiction.

$6\nmid 10$ since $10=2\cdot 5$ and $6=2\cdot 3$ ; the factor $3$ is absent from $10$ . Hence no surjective homomorphism $G\to G'$ can exist.

\boxed{\;\text{No homomorphism of }G\text{ onto }G'\text{ exists.}\;}