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Question:Let $G$ be a group and $N$ a normal subgroup of G, and $N\cong \mathbb{Z}$, $G/N\cong \mathbb{Z}/3\mathbb{Z}$. Show that G is abelian and is isomorphic to $\mathbb{Z}$ or $\mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}$.

I tried to solve this question by group action. Does $G/N$ acts on G? Is there someone who helps me?

Cave Johnson
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masutarou
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2 Answers2

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Let $b$ be a generator of $N$ and let $a$ be a representative in $G$ of the generator of $G/N$. Since $G/N \cong \mathbb{Z}/3\mathbb{Z}$, $a^3=b^k$ for some integer $k$ and since $N$ is normal in $G$, $$aba^{-1}=b^m$$ with some integer $m$. From $(aba^{-1})(aba^{-1})=ab^2a^{-1}=b^{2m}$,

$$b=a^3ba^{-3}=b^{m^3},$$

which follows $m=1$. Therefore $G$ is abelian.

For the strucure of $G$, it is sufficient to consider $a^3=b^k$ with $0 \leq k \leq 2$. If $k=0$, then $G\cong\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}$. If $k=1$, then $b$ is generated by $a$, i.e. $G\cong\mathbb{Z}$. For $k=2$, $c=a^2b^{-1}$ satisfies $c^3=b$, which follows $G\cong\mathbb{Z}$.

Tortoise Crane
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  • Abelianity

Consider an element $b$ outside $N$. Since $b^3 \in N$ and $N$ is abelian the map $x \mapsto bxb^{-1}$ is an automorphism of $N$ whose order divides $3$. Since Aut($\mathbb{Z}$) = $\mathbb{Z}_2$, conjugation must be trivial, i.e. group must be abelian.

Ivan Di Liberti
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