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I tried to show that $\mathbb Q$ with addition is not a cyclic group. Here is the proof:

If possible assume that there is $a\in \mathbb Q$ such that $\mathbb Q=\langle a \rangle$. Note that $a\neq 0$. For $1\in \mathbb Q$ we must have $1=ma$ for some $m\in \mathbb Z$.

Now for each $n\in \mathbb N$ we must have $\frac {1}{n}=m_n a$ for some $m_n\in \mathbb Z$. Then $1=nm_n a$ for each $n\in \mathbb N$. Then $ma=nm_n a$. Then we get $nm_n-m=0$. So $n$ must divide $m$ for each $n\in \mathbb N$. Then $m$ must be a zero which is not possible. So our supposition is wrong. Hence $\mathbb Q$ is not cyclic.

Is this correct?

Shaun
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Shreya Chauhan
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  • @perpetuallyconfused there are so many proofs available for this question. I wanted to prove it just using the definition of cyclic group not any other tool. So I wanted to check whether it is a right way or not. – Shreya Chauhan Nov 10 '20 at 03:07
  • I would remove 'then m must be a zero'. Just say that the previous sentence is a contradiction. – DanLewis3264 Nov 10 '20 at 03:15
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    This seems basically correct, but I'd use a shorter argument. If there's some rational number $a$ such that $\mathbb{Q} = \langle a \rangle$, then every rational number can be written as $na$ for some integer $n$. But there are obviously rationals that can't be written this way, for example $a/2$. – Connor Harris Nov 10 '20 at 03:16

2 Answers2

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You can show this more quickly in the following way: We must have $m_{1}a=a/2$ for some $m_{1} \in \mathbb{Z}$ so that $2m_{1}a=a \implies 2m_{1} = 1 \implies m_{1} = \frac{1}{2}$ but $\frac{1}{2} \notin \mathbb{Z}$.

Derek Luna
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Yes, the proof is correct.

There is another way: write $a=\frac pq$ with $\gcd(p,q)=1$. Then $\frac p{2q}\in\mathbb Q$ implies $\frac p{2q}=\frac{np}q$ for some $n\in\mathbb Z$, which means $2\mid1$, which is impossible.

There is yet another fun way:

Consider a surjective group homomorphism $f:A\rightarrow \mathbb Z$. For any such group homomorphism there is a group homomorphism $g:\mathbb Z\rightarrow A$ such that $f\circ g=id_{\mathbb Z}$.

But this question's answer constructs examples of surjective homomorphisms $f:A\rightarrow \mathbb Q$ such that no $g:\mathbb Q\rightarrow A$ can exist so that $f\circ g=id_{\mathbb Q}$. This means $\mathbb Q$ is not isomorphic to $\mathbb Z$.

Hope this helps.

awllower
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