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can someone help me out on this question. I am unsure how to proceed any further here. I have addedd my solution below, what I am unsure is the if truly $\mid \alpha \mid = 30$. We know that $\mid \phi(\alpha) = 1 \mid = 12$ but the other part i am unsure. Thank you.

First, we try to find if any onto homomorphisms exists. Assume $\phi : \mathbb{Z}_{30}$ onto $\mathbb{Z}_{12}$ is a homomorphism. Recall some properties given by Theorem 10.1 : Let $\phi$ be a homomorphism from a group $G$ to a group $\bar{G}$ and let $g$ be an element of $G$. Then if $\mid g \mid$ is finite, then $\mid \phi(g) \mid$ divides $\mid g \mid$.

Notice that since $\phi: \mathbb{Z}_{30}$ onto $\mathbb{Z}_{12}$ is an onto homorophism, this means that $\forall \alpha \in \mathbb{Z}_{30} \ \exists \beta \in \mathbb{Z}_{12} : \phi(\alpha) = \beta$. Let $\phi(\alpha)=1$, Notice here we have $\mid \phi(\alpha) = 1 \mid = 12$. Similarly $\mid \alpha \mid = 30$. By property 3, $12$ must divide $30$ which is not possible. Hence their is no onto homomorphism.

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  • $\begingroup$ Can I use First Isomorphism Theorem to answer your question? $\endgroup$ – Alan Wang Nov 28 '16 at 10:53
  • $\begingroup$ Yes please. But is my solution incorrect? $\endgroup$ – Emad Zamout Nov 28 '16 at 10:54
  • $\begingroup$ I think the statement $|\phi(\alpha)=1|=12$ is not clear. $\endgroup$ – Alan Wang Nov 28 '16 at 10:57
  • $\begingroup$ Your right the wording is really bad. the order of element 1 in $\mathbb{Z}_{12}$ is 1 $\endgroup$ – Emad Zamout Nov 28 '16 at 10:58
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Suppose there exists an homomorphism $\phi$ from $\Bbb{Z}_{30}$ onto $\Bbb{Z}_{12}$. Let $K=\ker\phi$. By First Isomorphism Theorem, $$\mathbb{Z}_{30}/K\cong \mathbb{Z}_{12}$$ This means that $30=12|K|$ which is a contradiction since there is no integral solution for $|K|$.

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  • $\begingroup$ Man that is much simpler than my approach at it. Thank you $\endgroup$ – Emad Zamout Nov 28 '16 at 11:06
  • $\begingroup$ Can you explain why their is no integral solutions for the kernal? $\endgroup$ – Emad Zamout Nov 28 '16 at 11:07
  • $\begingroup$ @EmadZamout $|K|=\frac{30}{12}=\frac{5}{2}$ which is impossible since $|K|$ must be a positive integer. $\endgroup$ – Alan Wang Nov 28 '16 at 11:37
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Let $a \in \mathbb Z_{30}$ and $b=\phi(a)$. Then $30b=30\phi(a)=\phi(30a)=\phi(0)=0$.

Also, $12b=0$ because $b \in \mathbb Z_{12}$. Therefore, $6b=0$ since $\gcd(30,12)=6$.

Therefore, every element of $\operatorname{im}\phi$ has order at most $6$ and so no element of $\operatorname{im}\phi$ has order $12$. Hence $\phi$ is never surjective because $\mathbb Z_{12}$ is cyclic of order $12$.

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