# 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.

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Great comment. I feel like so much freedom now with building function. But if it's not too much to ask, could some people please give me their honest opinion on how this test was structured. Solve 4 out of 6 with 1 for extra credit. Not only structure but emphasis on the questions and how well the solutions were written. – Eidbanger Oct 29 '11 at 8:13
I suggest that if you have a problem with a test given at your institution that in the first place you discuss it with your instructor and, if, after due consideration of your instructor's replies, you are still dissatisfied, you speak to the head of the department (or the head of undergraduate teaching if the department has one, or whatever). I think that's much better than trying to air any grievances in public. – Gerry Myerson Oct 29 '11 at 9:10
Well I would rather get some advice/ guidance if the test given was in fact not that good. – Eidbanger Oct 29 '11 at 15:59