$G$ is cyclic group. $\varphi:G\to H$ is homomorphism.
How do I show that $\ker \varphi$ and $\text{Im}\varphi$ are cyclic groups?
Thank you!
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5$\begingroup$ $\ker(\varphi) < G$, so it is cyclic. $Im(\varphi) = \langle \varphi(a)\rangle$ where $a$ is a generator of $G$ $\endgroup$– Prahlad VaidyanathanDec 1, 2013 at 11:34
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$\begingroup$ What is the meaning of $\ker\varphi <G$? Thank you! $\endgroup$– CS1Dec 1, 2013 at 11:35
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1$\begingroup$ It means $\ker(\varphi)$ is a subgroup of $G$. Do you know that a subgroup of a cyclic group must be cyclic? $\endgroup$– Prahlad VaidyanathanDec 1, 2013 at 11:38
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$\begingroup$ I don't remember if we learn it, but it's look familiar. It's hard to prove it? BTW, there is another way to prove it? Thank you! $\endgroup$– CS1Dec 1, 2013 at 11:57
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$\begingroup$ @PrahladVaidyanathan - can you write me a simple proof of it? thank you! $\endgroup$– CS1Dec 1, 2013 at 20:28
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2 Answers
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You can see the book for solving the second statement of your problem (see Theorem $13.9$).
By Theorem $13.3$ of the book on that link, ker$\phi$ is a normal subgroup of $G$ (In fact we only need ker$\phi$ is a subgroup of $G$) . Since $G$ is cyclic, and by the fact that subgroup of cyclic group is cyclic, we are done the first statement.
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$\begingroup$ Every subgroup of a cyclic group is normal... $\endgroup$– user1729Dec 18, 2013 at 16:15
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$\begingroup$ @user1729: You can check the link math.stackexchange.com/questions/295564/… $\endgroup$– TruongDec 18, 2013 at 16:19
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$\begingroup$ I understand the proof, I just don't understand your point(s). The fact that the kernel is normal is irrelevant - are you just saying that it is a subgroup? I also don't understand why you linked to a book, in the sense that this question isn't overly difficult so a standard answer or hint would be better... $\endgroup$– user1729Dec 18, 2013 at 16:28
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1$\begingroup$ It would be more self contained to give more information about what is said in Theorem 13.9. Links can go stale, and that would render this answer less useful. $\endgroup$– robjohn ♦Dec 19, 2013 at 1:53
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Hint: Note that $\ker(\phi)$ is a subgroup of $G$, so it is cyclic. Moreover, $\operatorname{im}(\phi)=\langle \phi(a)\rangle$ where $a$ is a generator of $G$.
(I have made this a community wiki answer, as it is simply Prahlad Vaidyanathan's comment. Answering this question removes it from the unanswered queue, and I dislike the other answer because it basically points at a book.)