UPSC 2024 Maths Optional Paper 2 Q2b — Step-by-Step Solution

15 marks · Section A

Group homomorphisms: kernel, image · Algebra · asked 4× in 13 yrs · Read the full method →

Question

Show that every homomorphic image of an abelian group is abelian, but the converse is not necessarily true.

Technique

Element-by-element verification using commutativity of $G$ ; sign homomorphism on $S_3$ as a counterexample for the converse.

Solution

Part 1 — Homomorphic image of abelian is abelian

Let $\phi:G\to H$ be a group homomorphism with $G$ abelian. Let $h_1,h_2\in\operatorname{Im}\phi$ ; pick $g_i\in G$ with $\phi(g_i)=h_i$ . Then

h_1 h_2=\phi(g_1)\phi(g_2)=\phi(g_1 g_2)\stackrel{G\text{ abelian}}{=}\phi(g_2 g_1)=\phi(g_2)\phi(g_1)=h_2 h_1.

So $\operatorname{Im}\phi$ is abelian. $\square$

Part 2 — Converse fails

Counterexample. Let $G=S_3$ (the symmetric group on 3 elements, which is non-abelian). Define the sign homomorphism $\phi:S_3\to\mathbb Z_2=\{+1,-1\}$ by

\phi(\sigma)=\operatorname{sgn}(\sigma).

This is a homomorphism with image $\mathbb Z_2$ , which is abelian. Yet $G=S_3$ is non-abelian (e.g., $(12)(13)\ne(13)(12)$ ).

The image $\operatorname{Im}\phi=\mathbb Z_2$ is abelian, but the domain $G=S_3$ is not. The converse fails.

Answer

\boxed{\;\text{Every homomorphic image of an abelian group is abelian; converse fails (e.g. }S_3\xrightarrow{\operatorname{sgn}}\mathbb Z_2\text{).}\;}