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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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    Just want to add, that m$y$ presence here is to learn in a right way, by myself, and not have to worry about silly teacher misguidance. (which is still minor, mind me) I love getting input and providing some myself. And this is the best place I can do it.2011-10-29

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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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    Well I would rather get some advice/ guidance if the test given was in fact not that good.2011-10-29