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Let $G$ be gorup and $H$ be normal subgroup of $G$.

Suppose that $H \cong \mathbb{Z}$ ($\mathbb{Z}$:integer) and $G/H \cong \mathbb{Z}/ n \mathbb{Z}　(\mathbb{Z} \ni n \geq 2 )$.

Then I'm stuck in next problems.

(1)If $n$ is odd number, G is abelian group.

(2)Classfy the group G up to isomorphism

Please tell me any idea and help me.

2 Answers 2

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Let me give you a headstart: in general, if $N \unlhd G$, then conjugation on $N$ induces an automorphism of $N$, hence $G/C_G(N) \hookrightarrow Aut(N)$. Here $C_G(N)=\{g \in G: g^{-1}ng=n, $ for all $n \in N\}$, the centralizer of $N$ in $G$. Now $Aut(\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$. Hence, if $n$ is odd, we see that in your case $G/C_G(H)$ must be trivial, that is $H \subseteq Z(G)$. Since $G/H$ is cyclic, it follows that also $G/Z(G)$ is cyclic and it is well known that implies that $G$ is abelian. This proves (1). Can you work on (2) now?

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    I read your response and I don't understand where "assumption $n$:odd number" uses. I have tried it,but it dosen't work. If you do not mind, please give me help about (2) as (1). – 2017-01-21
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Hint: Classify the homomorphisms $\Bbb{Z}/n\Bbb{Z}\ \longrightarrow\ \operatorname{Aut}(\Bbb{Z})$.

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    Sorry,What is definition of that homomorphism? – 2017-01-21
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    I didn't define a homomorphism. I suggested you figure out what *all* homomorphisms from $\Bbb{Z}/n\Bbb{Z}$ to $\operatorname{Aut}(\Bbb{Z})$ are. Are you familiar with semidirect products? – 2017-01-21
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    No,I can't figure out. – 2017-01-21
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    I am unfamiliar with semidirect products. – 2017-01-21
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    I honestly have no idea how to answer this question without semidirect products. – 2017-01-21
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    Please tell me the idea using semidirect products – 2017-01-21
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    When I ascertain it, I understand semidirect products. – 2017-01-21