# Thread: [SOLVED] Proper Subgroups of Q

1. ## [SOLVED] Proper Subgroups of Q

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

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

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

2. Originally Posted by Haven
Prove that $\displaystyle \mathbb{Q}$ is not isomorphic to a proper subgroup of itself.

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

I know there's a contradiction here, but I can't see it
Assume G be a proper subgroup of $\displaystyle (\mathbb{Q},+)$ such that $\displaystyle \phi : \mathbb{Q} \rightarrow G$ is a bijection.
Let $\displaystyle \frac{p}{q} \in (\mathbb{Q} \setminus G)$. Since $\displaystyle \phi$ is a bijection, if $\displaystyle x = \frac{p/q}{\phi(1)}$, then $\displaystyle \phi(x)=\frac{p}{q}$, where $\displaystyle \phi(x)=x\phi(1)$ and $\displaystyle \phi(1) \neq 0$. It contradicts that $\displaystyle \frac{p}{q}$ is not a member of G. Thus $\displaystyle (\mathbb{Q}, +)$ is not isomorphic to a proper subgroup of itself.

3. Originally Posted by aliceinwonderland
Assume G be a proper subgroup of $\displaystyle (\mathbb{Q},+)$ such that $\displaystyle \phi : \mathbb{Q} \rightarrow G$ is a bijection.
Let $\displaystyle \frac{p}{q} \in (\mathbb{Q} \setminus G)$. Since $\displaystyle \phi$ is a bijection, if $\displaystyle x = \frac{p/q}{\phi(1)}$, then $\displaystyle \phi(x)=\frac{p}{q}$, where $\displaystyle \phi(x)=x\phi(1)$ and $\displaystyle \phi(1) \neq 0$. It contradicts that $\displaystyle \frac{p}{q}$ is not a member of G. Thus $\displaystyle (\mathbb{Q}, +)$ is not isomorphic to a proper subgroup of itself.
I do not believe this is a proof. You assume $\displaystyle \phi(x) = x\phi(1)$, 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. $\displaystyle \mathbb{Z} \times \mathbb{Z} \times \ldots \cong <1> \times \mathbb{Z} \times \ldots$.

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

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

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

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

Or something along those lines...

4. Originally Posted by Swlabr
I do not believe this is a proof. You assume $\displaystyle \phi(x) = x\phi(1)$, 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. $\displaystyle \mathbb{Z} \times \mathbb{Z} \times \ldots \cong <1> \times \mathbb{Z} \times \ldots$.

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

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

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

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

Or something along those lines...
I do not believe this is false. If $\displaystyle \phi$ is an isomorphism between (Q, +) and its subgroup, then $\displaystyle \phi(x)=x\phi(1)$ 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).
$\displaystyle \phi (\frac{p}{q})=\phi(\frac{1}{q} + ... +\frac{1}{q})=p \phi(\frac{1}{q})$.
$\displaystyle \phi (1)=\phi(\frac{q}{q})=q\phi(\frac{1}{q})$. It follows that $\displaystyle \phi(\frac{1}{q})=\frac{\phi(1)}{q}$.
Then, $\displaystyle \phi(x)=\phi(\frac{p}{q})=\frac{p}{q}(\phi(1))=x\p hi(1)$.

5. Originally Posted by aliceinwonderland
I do not believe this is false. If $\displaystyle \phi$ is an isomorphism between (Q, +) and its subgroup, then $\displaystyle \phi(x)=x\phi(1)$ 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).
$\displaystyle \phi (\frac{p}{q})=\phi(\frac{1}{q} + ... +\frac{1}{q})=p \phi(\frac{1}{q})$.
$\displaystyle \phi (1)=\phi(\frac{q}{q})=q\phi(\frac{1}{q})$. It follows that $\displaystyle \phi(\frac{1}{q})=\frac{\phi(1)}{q}$.
Then, $\displaystyle \phi(x)=\phi(\frac{p}{q})=\frac{p}{q}(\phi(1))=x\p hi(1)$.
I apologise for any ambiguity in my previous post. I was trying to say the claim required proof, not that it was wrong.