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$G$ 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!

CS1
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    $\ker(\varphi) < G$, so it is cyclic. $Im(\varphi) = \langle \varphi(a)\rangle$ where $a$ is a generator of $G$ – Prahlad Vaidyanathan Dec 01 '13 at 11:34
  • What is the meaning of $\ker\varphi <G$? Thank you! – CS1 Dec 01 '13 at 11:35
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    It means $\ker(\varphi)$ is a subgroup of $G$. Do you know that a subgroup of a cyclic group must be cyclic? – Prahlad Vaidyanathan Dec 01 '13 at 11:38
  • 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! – CS1 Dec 01 '13 at 11:57
  • @PrahladVaidyanathan - can you write me a simple proof of it? thank you! – CS1 Dec 01 '13 at 20:28
  • The proof will be there in pretty much any book on Group theory (for instance, look at Gallian/Herstein/Artin/etc) – Prahlad Vaidyanathan Dec 02 '13 at 07:01
  • You can check http://answers.yahoo.com/question/index?qid=20101121123720AAmzoBX – Truong Dec 18 '13 at 15:41
  • Concerning every subgroup of a cyclic group is cyclic. Say $G=\left< x\right>$. Let $H<G$. If $|H|=1$ then $H=\left< e\right>$. Let $|H|>1$. Let $n$ be the least positive integer such that $x^n \in H$. Let $y \in H$ so $y=x^m$ for some $m \in \mathbb{Z}$. By the division algorithm, there exists unique $q,r \in \mathbb{Z}$ such that $m=nq+r$ and $0\leq r<n$. Hence, $y=x^m=x^{nq+r}=x^{nq}x^r$ and so $x^m*x^{-nq}=x^r$ but $x^m,x^{-nq} \in H$ so $x^r\in H$ but $n$ is the least positive integer such that $x^n \in H$, so as $r<n$, it must be $r=0$ so $y=x^{nq}\in H$ so $H=\left<x^n\right>$. –  Dec 18 '13 at 16:39

2 Answers2

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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.

Truong
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  • Every subgroup of a cyclic group is normal... – user1729 Dec 18 '13 at 16:15
  • @user1729: I edited my answer – Truong Dec 18 '13 at 16:18
  • @user1729: You can check the link http://math.stackexchange.com/questions/295564/a-subgroup-of-a-cyclic-group-is-cyclic-understanding-proof – Truong Dec 18 '13 at 16:19
  • 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... – user1729 Dec 18 '13 at 16:28
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    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. – robjohn Dec 19 '13 at 01:53
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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.)

user1729
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