# True or false. Order of $\phi (a)$ is equal to the order of $a$ [duplicate]

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?

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

• Consider the trivial homomorphsim $\phi(a) = 1_{G_2}$ – reuns Nov 4 '16 at 20:02
• I would like to know if my counter example is correct. – combo student Nov 4 '16 at 20:02
• @Dietrich: It is certainly not a duplicate, since it (implicitly) asks about a specific argument different from the one in the earlier question. – Brian M. Scott Nov 4 '16 at 20:03
• @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. – Dietrich Burde Nov 4 '16 at 20:05
• @Dietrich: No, it isn’t, and providing a correct argument does not answer the OP’s real question. – Brian M. Scott Nov 4 '16 at 20:06

Your map isn’t well-defined: $_{15}=_{15}$, so $\varphi(_{15})$ should be equal to $\varphi(_{15})$, but in fact $_6\ne_6=_6$.
HINT: Take $G_1$ to be any group with at least two elements and $G_2$ to be the trivial (one-element)group.
What you wrote down is not a homomorphism. If it were then: $$_6=\varphi(_{15}) = \varphi(_{15} + _{15}) = \varphi(_{15}) + \varphi(_{15}) = _6 + _6 = _6.$$
• Thank you. What if i change it to $\phi([a]_{15}) = [3a]_6$? – combo student Nov 4 '16 at 20:07
• @combostudent: That doesn’t work either: $_{15}=_{15}$, but $_6\ne_6=_6$, so this map also is not well-defined. – Brian M. Scott Nov 4 '16 at 20:10
• where are you getting $48 mod 6$ ? – combo student Nov 4 '16 at 20:14