(January 2012) Let \(G\) and \(H\) be groups. Recall that a homomorphism \(\varphi:G\to H\) is said to be trivial if \(\varphi(g)=e_H\) for all \(g\in G\). If \(|G|=144\) and \(|H|=25\), prove that any homomorphism \(\varphi:G\to H\) is trivial.
Solution
Proof. Giveng∈G. By Lagrange’s Theorem for G, we have g144 =eG Since φ is a homomorphism, we have (φ(g))144 =φ(g144) =φ(eG) =eH. Thus, φ(g), which is an element of H, has finite order m = o(φ(g)) that divides 144. On the other hand, by Lagrange’s Theorem forH, we also havem|25, since|H| = 25. Thus, m is a common divisor of 144 = 2432 and 25 = 52. Since gcd(144, 25) = gcd(2432, 52) = 1, we must have o ( φ(g) ) = 1. Thus, φ(g) =eH, as desired. QED