True or false. (Prove or give a counterexample.) Let $G$ be a group. Then $|g| = |\phi(g)|$, for all homomorphisms $\phi: G \to G$ and all $g \in G$.
Solution. False. $\phi: \mathbb Z_{10} \to \mathbb Z_{12}$ defined by $\phi(x)=0$ for all $x \in \mathbb Z_{10}$ is a counterexample. This function is a homomorphism because $\phi(x+y) = 0 = 0+0 = \phi(x) + \phi(y)$ for all $x, y \in \mathbb Z$ (this function is discussed in problem 3 in assignment 7 as the function sending $[1]$ to $[0]$). The order of $g=1$ is infinity. The order of $\phi(1)=0$ is one.
The problem is that he said $G \to G$ is equivalent to $\mathbb Z_{10} \to \mathbb Z_{12}$, which is think is not right. Explain please?
EDIT: Please look at this test and give me your honest opinion, http://zimmer.csufresno.edu/~ovega/teaching/151/Exam2Solutions.pdf