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Consider the following statements:

I. There exists a proper subgroup $G$ of $(\mathbb{Q}, +)$ such that $\mathbb{Q}/G$ is a finite group.

II. There exists a subgroup $G$ of $(\mathbb{Q}, +)$ such that $\mathbb{Q}/G$ is isomorphic to $(\mathbb{Z}, +)$.

The above question was asked in $\text{GATE 2024}$.

Statement (I) is not true as $(\mathbb{Q}, +)$ has no proper subgroup of finite index. (I have a proper proof of this statement by way of contradiction.)

For statement (II), I found from site search that this statement is also not true. Reason is simply given that $(\mathbb{Q}, +)$ is a divisible group and hence every quotient group of $(\mathbb{Q}, +)$ is divisible. Therefore, there does not exists a subgroup $G$ of $(\mathbb{Q}, +)$ such that $\mathbb{Q}/G$ is isomorphic to $(\mathbb{Z}, +)$.

Question: Is there any other method to show that the statement (II) is not true, which is independent of divisibility. Because my concepts of divisibility of groups are weak.

I tried to think in this way: Since each non identity element of $(\mathbb{Z}, +)$ is of infinite order and my intuition says that each element of a quotient group of $(\mathbb{Q}, +)$ is of finite order, therefore both can not be isomorphic. I am not sure that my intuition is correct and if it is true then how can I prove the above intuition.

Kindly help me.

Sammy Black
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Sandeep Tiwari
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1 Answers1

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I am writing the answer using the link provided by @ghc1997.
Suppose that there were an isomorphism $f : \mathbb{Q}/G\to \mathbb{Z}$. Then there exists some $q +G\in \mathbb{Q}/G$ such that $f(q+G) = 1$. But then, $x = f(q/2+G)$ is an integer satisfying $$2x = x+x = f(q/2+G) + f(q/2+G) = f(q/2 + q/2+G) = f(q+G)= 1,$$ which is impossible. Therefore there is no such $f$.

Sandeep Tiwari
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