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I was reading an example about describing homomorphisms and I'm having a bit of trouble with this one.

Show there is no group homomorphism $f : \mathbb{Z}_{10} \to \mathbb{Z}_{25}$ such that $f(1) = 3$. The first step of the proof is $|1| = 10$ in $\mathbb{Z}_{10}$, so $|f(1)| \in\{ 1, 2, 5, 10\}$.

Does this mean that $|f(1)|$ can be $1$, $2$, $5$ or $10$? I see that $|1| = 10$ since $1$ is a generator of $\mathbb{Z}_{10}$, and that $|f(1)|$ seems to be a factor of $10$. If $f(1)$ is the image of $1$ under $f$, how would I find the order of $f(1)$?

Thanks for your help.

user2097
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jstnchng
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1 Answers1

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If $x^n=e$ and $\phi$ a homomorphism, then $\phi(x^n)=(\phi(x))^n$ is the identity element too. Therefore, the order of $\phi(x)$ divides $n$, which gives a solution to present problem.

user2097
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