Answer to Question #248053 in Abstract Algebra for Endalew Erdaw

Question

Prove that Z27 is not a homomorphic image of Z72

Answer 1

Let us prove that $\Z_{27}$ is not a homomorphic image of $\Z_{72}$ using the method by contradiction. Suppose that $\Z_{27}$ is a homomorphic image of $\Z_{72}$ under some homomorphism $\varphi.$ Let $a$ be a generator of $\Z_{72}.$ Then the order of $a$ is equal to 72. Taking into account that the order of $\varphi(a)$ divides the order of $a,$ we conclude that the order of $\varphi(a)$ divides 72. Since $\Z_{27}$ is a homomorphic image of

$\Z_{72},$ we conclude that $\varphi(a)$ is a generator of $\Z_{27}$, and hence $\varphi(a)$ is of order 27. Since 27 does not divide 72, we get a contradiction. Therefore, $\Z_{27}$ is not a homomorphic image of $\Z_{72}.$