Group theory $G \triangleright H , H \cong \mathbb{Z}, G/H \cong \mathbb{Z}/ n \mathbb{Z}$.Then Classfication of $G$

Question

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.

Answer 1

Hint: Classify the homomorphisms $\Bbb{Z}/n\Bbb{Z}\ \longrightarrow\ \operatorname{Aut}(\Bbb{Z})$.

Answer 2

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?