← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q2a — Step-by-Step Solution

10 marks · Section A

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

Question

If $G$ and $H$ are finite groups whose orders are relatively prime, then prove that there is only one homomorphism from $G$ to $H$, the trivial one.

Technique

First Isomorphism Theorem plus Lagrange: $|\operatorname{im}\varphi|$ divides both $|G|$ and $|H|$, hence divides their gcd $=1$.

Solution

Let $\varphi:G\to H$ be a homomorphism, with $\gcd(|G|,|H|)=1$.

Step 1 — The image is a subgroup of $H$

$\varphi(G)$ is a subgroup of $H$ (image of a homomorphism). By Lagrange’s theorem,

Step 2 — The image’s order also divides $|G|$

By the First Isomorphism Theorem, $\varphi(G)\cong G/\ker\varphi$, so

Thus $|\varphi(G)|$ is a common divisor of $|G|$ and $|H|$:

Step 3 — Conclude

Therefore $|\varphi(G)|=1$, i.e. $\varphi(G)=\{e_H\}$. So $\varphi(g)=e_H$ for every $g\in G$: $\varphi$ is the trivial homomorphism. As $\varphi$ was an arbitrary homomorphism, the trivial map is the only homomorphism $G\to H$.

Answer