Exercise 10, Section 1.3 of Hungerford’s Abstract Algebra

Question

An infinite group is cyclic if and only if it is isomorphic to each of its proper subgroups.

My attempt: Let $G$ be an infinite group. $(\Rightarrow)$ Suppose $G=\langle a\rangle$. Let $H$ be a proper subgroup of $G$, i.e. $H\neq \{e\}, G$. By theorem 5 section 1.3, $H=\langle a^m\rangle$ where $m$ is least positive integer such that $a^m\in H$. Since $G$ is infinite, we have $a^i\neq a^j$, if $ i\neq j$. So $H$ is infinite. By theorem 2 section 1.3, $G\cong \Bbb{Z}$ and $H\cong \Bbb{Z}$. Since $\cong$ is an equivalence relation, we have $G\cong H$.

$(\Leftarrow)$ Suppose $G\cong H$, for all $\{e\}\neq H\lt G$. Let $a\in G\setminus \{e\}$. By hypothesis, $G\cong \langle a\rangle$. That is $\exists f:\langle a\rangle \to G$ such that $f$ is homomorphism and bijective. By theorem 5 section 1.3, $\text{Im} f=\langle f(a)\rangle =G$. Thus $G$ is cyclic. Is my proof correct?

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Edit: Thanks to User Greg Martin for commenting, possibility of $\langle a\rangle$ not being a proper subgroup. If $\langle a\rangle$ is not a proper subgroup, then $\langle a\rangle =\{e\}$ or $\langle a\rangle =G$. Since $e\neq a\in \langle a\rangle$, we have $\langle a\rangle \neq \{e\}$. So $\langle a\rangle =G$. Thus $G$ is cyclic.

Answer 1

Your direct part is correct. For the reverse part of the equivalence, I would say the following.

Let $a \in G \setminus \{e\}$. If $G = \langle a\rangle$, we are done. Otherwise $\langle a \rangle$ is a proper subgroup of $G$ and isomorphic to $G$ by hypothesis. $\langle a \rangle$ is infinite, as $G$ is supposed to be infinite. According to theorem 2, section 1.3, $\langle a \rangle$ is isomorphic to the additive group $\mathbb Z$ and therefore $G$ too. We therefore get the desired conclusion that $G$ is cyclic infinite.