[SOLVED] Proper Subgroups of Q

**Haven** · February 5th 2010, 12:49 PM

Prove that $\mathbb{Q}$ is not isomorphic to a proper subgroup of itself.

So, far i have:
Let $G \leq (\mathbb{Q},+)$
Let $\phi : G \rightarrow (\mathbb{Q},+)$
Let $\frac{p}{q}$ not in $G$
Since $\phi$ is bijective $\phi(x) = \frac{p}{q}$ for some $x \in G$

I know there's a contradiction here, but I can't see it

**aliceinwonderland** · February 5th 2010, 09:40 PM

Originally Posted by Haven

Prove that $\mathbb{Q}$ is not isomorphic to a proper subgroup of itself.

So, far i have:
Let $G \leq (\mathbb{Q},+)$
Let $\phi : G \rightarrow (\mathbb{Q},+)$
Let $\frac{p}{q}$ not in $G$
Since $\phi$ is bijective $\phi(x) = \frac{p}{q}$ for some $x \in G$

I know there's a contradiction here, but I can't see it

Assume G be a proper subgroup of $(\mathbb{Q},+)$ such that $\phi : \mathbb{Q} \rightarrow G$ is a bijection.
Let $\frac{p}{q} \in (\mathbb{Q} \setminus G)$ . Since $\phi$ is a bijection, if $x = \frac{p/q}{\phi(1)}$ , then $\phi(x)=\frac{p}{q}$ , where $\phi(x)=x\phi(1)$ and $\phi(1) \neq 0$ . It contradicts that $\frac{p}{q}$ is not a member of G. Thus $(\mathbb{Q}, +)$ is not isomorphic to a proper subgroup of itself.

**Swlabr** · February 5th 2010, 11:48 PM

Originally Posted by aliceinwonderland

Assume G be a proper subgroup of $(\mathbb{Q},+)$ such that $\phi : \mathbb{Q} \rightarrow G$ is a bijection.
Let $\frac{p}{q} \in (\mathbb{Q} \setminus G)$ . Since $\phi$ is a bijection, if $x = \frac{p/q}{\phi(1)}$ , then $\phi(x)=\frac{p}{q}$ , where $\phi(x)=x\phi(1)$ and $\phi(1) \neq 0$ . It contradicts that $\frac{p}{q}$ is not a member of G. Thus $(\mathbb{Q}, +)$ is not isomorphic to a proper subgroup of itself.

I do not believe this is a proof. You assume $\phi(x) = x\phi(1)$ , but this is not necessarily so. Essentially, you have produced a generic proof which says that a proper subgroup of a group cannot be isomorphic to the group itself. This is not so. An obvious counter-example is the infinite cartesian product of the integers. $\mathbb{Z} \times \mathbb{Z} \times \ldots \cong <1> \times \mathbb{Z} \times \ldots$ .

However, the result you used is true, but needs to be proven.

Notice that for $x \in \mathbb{Z}$ , $x\phi = x$ . So the result now just needs to be shown for the denominators.

$x\phi = 1/p \Rightarrow (px)\phi = 1 \Rightarrow x = 1/p$ as the kernel is trivial.

Thus, $\frac{p}{q} \phi = \frac{p\phi}{q\phi} = \frac{p}{q}$ as required.

Or something along those lines...

**aliceinwonderland** · February 6th 2010, 11:05 AM

Originally Posted by Swlabr

I do not believe this is a proof. You assume $\phi(x) = x\phi(1)$ , but this is not necessarily so. Essentially, you have produced a generic proof which says that a proper subgroup of a group cannot be isomorphic to the group itself. This is not so. An obvious counter-example is the infinite cartesian product of the integers. $\mathbb{Z} \times \mathbb{Z} \times \ldots \cong <1> \times \mathbb{Z} \times \ldots$ .

However, the result you used is true, but needs to be proven.

Notice that for $x \in \mathbb{Z}$ , $x\phi = x$ . So the result now just needs to be shown for the denominators.

$x\phi = 1/p \Rightarrow (px)\phi = 1 \Rightarrow x = 1/p$ as the kernel is trivial.

Thus, $\frac{p}{q} \phi = \frac{p\phi}{q\phi} = \frac{p}{q}$ as required.

Or something along those lines...

I do not believe this is false. If $\phi$ is an isomorphism between (Q, +) and its subgroup, then $\phi(x)=x\phi(1)$ must hold. Here is why.

Let x be an arbitrary member in Q, let's say
x= p/q (if x=0, it is trivial).
$\phi (\frac{p}{q})=\phi(\frac{1}{q} + ... +\frac{1}{q})=p \phi(\frac{1}{q})$ .
$\phi (1)=\phi(\frac{q}{q})=q\phi(\frac{1}{q})$ . It follows that $\phi(\frac{1}{q})=\frac{\phi(1)}{q}$ .
Then, $\phi(x)=\phi(\frac{p}{q})=\frac{p}{q}(\phi(1))=x\p hi(1)$ .

**Swlabr** · February 7th 2010, 01:17 AM

Originally Posted by aliceinwonderland

I do not believe this is false. If $\phi$ is an isomorphism between (Q, +) and its subgroup, then $\phi(x)=x\phi(1)$ must hold. Here is why.

Let x be an arbitrary member in Q, let's say
x= p/q (if x=0, it is trivial).
$\phi (\frac{p}{q})=\phi(\frac{1}{q} + ... +\frac{1}{q})=p \phi(\frac{1}{q})$ .
$\phi (1)=\phi(\frac{q}{q})=q\phi(\frac{1}{q})$ . It follows that $\phi(\frac{1}{q})=\frac{\phi(1)}{q}$ .
Then, $\phi(x)=\phi(\frac{p}{q})=\frac{p}{q}(\phi(1))=x\p hi(1)$ .

I apologise for any ambiguity in my previous post. I was trying to say the claim required proof, not that it was wrong.

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