# Is it true that $|g| = |\phi(g)|$ for all homomorphisms $\phi: G \to G$ and $g \in G$?

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

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Did you ask your teacher? What happens if you consider the morphism $\phi:\mathbb Z_{10}\to\mathbb Z_{10}$ defined by the same formula? – Did Oct 29 '11 at 7:46
I added the relevant question. The original source is here. (Note that the solution seems to have a lot of errors.) – Srivatsan Oct 29 '11 at 7:54
Could you please point on those errors? – Eidbanger Oct 29 '11 at 7:56
@Eidbanger To be frank, this question is not that difficult. Also, even if the official solution is wrong, it still has a correct approach that you must try to understand. Since these solutions are anyway provided *after* the test, this has certainly not impacted your performance in the test. You should point out the mistake to the teacher and ask what correct solution was intended, but I do not see a reason to feel cheated by the mistake. – Srivatsan Oct 29 '11 at 8:12
So, replace one $\mathbb Z_{12}$ and one $\mathbb Z$ by $\mathbb Z_{10}$ and everybody is happy. // Some remarks about wording and etiquette: (1) Typos are not the same as wrong answers. (2) The best way to approach him about it includes communicating the address of this webpage to him. – Did Oct 29 '11 at 8:14

The answer is right, but the solution is odd. The question asks for $G\to G$, the answer doesn't give that, and also the answer veers mid-course from the domain being ${\bf Z}_{10}$ to it being $\bf Z$.
Let $G$ be any group with more than one element, let $\phi$ map everything in $G$ to the identity element of $G$, end of story.