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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?

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.

user264745
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    Your proof looks good to me. Maybe, you should just add a sentence to explain why $\langle a \rangle$ can't be finite as you haven't used explicitly the hypothesis that $G$ is infinite. – mathcounterexamples.net Apr 04 '23 at 19:39
  • @mathcounterexamples.net you mean in $(\Leftarrow )$ part of proof, if yes, then I think we don’t care about cardinality of $\langle a\rangle$. We used $G$ is infinite assumption in showing existence of $a$ (non identity element of $G$). – user264745 Apr 04 '23 at 19:51
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    The second half is basically a correct proof of the statement "If a group with more than one element is isomorphic to each of its proper subgroups, then it is cyclic". (We don't know whether $\langle a\rangle$ is a proper subgroup or not, but if it's not then it's trivially isomorphic to $G$.) It doesn't use the infinite assumption, and indeed cyclic groups of prime order also have this property since there are no proper subgroups. – Greg Martin Apr 04 '23 at 20:16

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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.

mathcounterexamples.net
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