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If it's true write a proof. If it's false, give a counter example.

If $\phi : G_1 \rightarrow G_2$ is a homomorphism and $a\in G$ then the order of $\phi(a)$ then is equal to the order of $a$.

My attempt: This is false. Consider $\phi:Z_{15} \rightarrow Z_6$ difined by $\phi([a]_{15})$=$[a]_6$.

This is homomorphism since $\phi([a]_{15}+[b]_{15})= \phi([a+b]_{15})=[a+b]_6=[a]_6+[b_6]= \phi[a]_{15}+\phi[b]_{15}.$ Let $a\in Z_{15}=3$. The order of $3$ is $5$ since $3+3+3+3+3=0mod15$. The order of $\phi(3)$ is $2$ since $3+3=0mod6$.

EDIT: My counter example is not well-defined. What if i change it to $\phi([a]_{15}) = [3a]_6$ and follow the same steps?

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marked as duplicate by Dietrich Burde, Watson, suomynonA, Leucippus, Jack's wasted life Nov 5 '16 at 4:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Consider the trivial homomorphsim $\phi(a) = 1_{G_2}$ $\endgroup$ – reuns Nov 4 '16 at 20:02
  • $\begingroup$ I would like to know if my counter example is correct. $\endgroup$ – combo student Nov 4 '16 at 20:02
  • $\begingroup$ @Dietrich: It is certainly not a duplicate, since it (implicitly) asks about a specific argument different from the one in the earlier question. $\endgroup$ – Brian M. Scott Nov 4 '16 at 20:03
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    $\begingroup$ @BrianM.Scott But the discussion there about the homomorphism $\mathbb{Z}_{12}\rightarrow \mathbb{Z}_{10}$ is just too similar to the one given here; and anyway, the trivial homomorphism finishes it all. $\endgroup$ – Dietrich Burde Nov 4 '16 at 20:05
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    $\begingroup$ @Dietrich: No, it isn’t, and providing a correct argument does not answer the OP’s real question. $\endgroup$ – Brian M. Scott Nov 4 '16 at 20:06
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Your map isn’t well-defined: $[3]_{15}=[18]_{15}$, so $\varphi([3]_{15})$ should be equal to $\varphi([18]_{15})$, but in fact $[3]_6\ne[18]_6=[0]_6$.

HINT: Take $G_1$ to be any group with at least two elements and $G_2$ to be the trivial (one-element)group.

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What you wrote down is not a homomorphism. If it were then: $$[0]_6=\varphi([0]_{15}) = \varphi([5]_{15} + [10]_{15}) = \varphi([5]_{15}) + \varphi([10]_{15}) = [5]_6 + [10]_6 = [3]_6.$$

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  • $\begingroup$ Thank you. What if i change it to $\phi([a]_{15}) = [3a]_6$? $\endgroup$ – combo student Nov 4 '16 at 20:07
  • $\begingroup$ @combostudent: That doesn’t work either: $[1]_{15}=[16]_{15}$, but $[3]_6\ne[48]_6=[0]_6$, so this map also is not well-defined. $\endgroup$ – Brian M. Scott Nov 4 '16 at 20:10
  • $\begingroup$ where are you getting $48 mod 6$ ? $\endgroup$ – combo student Nov 4 '16 at 20:14
  • $\begingroup$ @combostudent I think checking out this question will prove instructive. $\endgroup$ – Ken Duna Nov 4 '16 at 20:14

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