# These groups are not isomorphic, any hints?

Please me some clues verifying the following groups are not isomorphic. I 've found this problem interesting.

1. $(\mathbb Z,+)$ and $(\mathbb Q,+)$
2. $(\mathbb Q,+)$ and $(\mathbb R,+)$
3. $(\mathbb Z[x],+)$ and $(\mathbb Q[x],+)$
4. $(\mathbb Q[x],+)$ and $(\mathbb Q,+)$

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Do you have any ideas for any of your questions? – Chris Eagle Oct 13 '12 at 9:52
Hints for the first two. 1) Is $(\mathbb{Q},+)$ cyclic?. 2) Is $\mathbb{R}$ countable?... – fretty Oct 13 '12 at 10:02

1. In $\Bbb Q$, $z+z=c$ is solvable for all $c$.
2. There is no bijection $\Bbb Q\to\Bbb R$.
3. same as 1. (for $c=1$, say)
4. Consider the image of $1$ of a homomorphism $f:\Bbb Q\to \Bbb Q[x]$.
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Thanks. I v'e just forgotten that Q is not cyclic. – Nancy Rutkowskie Oct 13 '12 at 10:06
I don't understand your answer for 4. How does knowing where $1$ is sent tell us where 48267/4672625 is sent say? I understand that it determines where all integers are sent, but fractions are a different matter (we are in an additive group). – fretty Oct 13 '12 at 10:08
@fretty: $f(48267/4672625)$ must have the property that, when you add together $4672625$ copies of it, you get $f(48267)$. How many things are there in $\mathbb{Q}[x]$ with this property? – Chris Eagle Oct 13 '12 at 10:22
Ah I was having a dense moment. – fretty Oct 13 '12 at 11:33

Although Berci's first nice hint is easy to understand, I found another way just for it. $\mathbb Q$ is a divisible group but $\mathbb Z$ is not.

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This is actually almost exactly the same idea. – Mikko Korhonen Oct 13 '12 at 10:39
Nice example, Babak – amWhy Mar 28 '13 at 0:35

For the 4th part:

$f : \mathbb{Q} \to \mathbb{Q}[X]$ is isomomorphism let $f(1)=a$ then $f(m)=f(1 + \dots + 1) = mf(1) = ma$ also $f(0)=0$ thus we get $f(1-1)=f(0)$ which implies $f(1)+f(-1)=0$ thus $f(-1)=-f(1)=-a$ then $f(p) = p*f(1)$ also $f(p)=f(q*\frac{p}{q})=q*f(\frac{p}{q})$ from this we can get $f(\frac{p}{q})=\frac{p}{q}*f(1)$ therefore for all $r \in \mathbb{Q}$, $f(r)=rf(1)$ then it can be said $f$ cannot be surjective.

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Thanks for the attempt. – Nancy Rutkowskie Oct 15 '12 at 15:39