# Thread: Group Isomorphism

1. ## Group Isomorphism

Prove that $(\mathbb{Z}[x],+) \cong (\mathbb{Q}^{\times},\cdot)$.

I am not seeing it.

2. Originally Posted by Gamma
Prove that $(\mathbb{Z}[x],+) \cong (\mathbb{Q}^{\times},\cdot)$.

I am not seeing it.
If $a\in \mathbb{Q}^{\times}$ we can write $a = \prod_{n=0}^{\infty} p_n^{e_n}$ where $e_n\in \mathbb{Z}$ and $p_1,p_2,p_3,...,$ are the primes in increasing order.
Thus, I think the natural map is $\phi \left( \sum_{n=0}^{\infty} a_n x^n \right) = \prod_{n=0}^{\infty} p_n^{e_n}$.

Of course, the infinite product is really finite and the infinite sum is really finite because
almost all terms are 1 in the infinite product and almost all terms are zero in the infinite sum.

Now of course $\phi \left( \sum_{n=0}^{\infty}a_n x^n + \sum_{n=0}^{\infty}b_n x^n \right) = \phi \left( \sum_{n=0}^{\infty} a_n x^n \right) \phi \left( \sum_{n=0}^{\infty} b_n x^n \right)$ since addition of exponents is adding the exponents.
This happens to be one-to-one and onto.

3. Originally Posted by ThePerfectHacker

Thus, I think the natural map is $\phi \left( \sum_{n=0}^{\infty} a_n x^n \right) = \prod_{n=0}^{\infty} p_n^{e_n}$.
the power of $p_n$ in the RHS should be $a_n$. so i think you meant $\phi \left( \sum_{n=0}^{m} a_n x^n \right) = \prod_{n=0}^{m} p_n^{a_n},$ which is indeed a group isomorphism.

4. Argh! I spent the whole night thinking of the problem, and now that I have found a solution, I see that ThePerfectHacker and NonCommAlg have got here before me and deprived me of the chance to show off.

Only kidding. I come here to help, to learn from others and to share what I know, not to show off.

Anyway, this is a very beautiful problem. The fact that every positive rational number can be written uniquely in the form $2^{a_1}3^{a_2}5^{a_3}\cdots$ where $a_i\in\mathbb Z$ is very beautiful indeed.

5. Originally Posted by NonCommAlg
the power of $p_n$ in the RHS should be $a_n$. so i think you meant $\phi \left( \sum_{n=0}^{m} a_n x^n \right) = \prod_{n=0}^{m} p_n^{a_n},$ which is indeed a group isomorphism.
Of course that is what I meant. Thank you for the correction.

6. By the way, just realized of something.

Originally Posted by Gamma
Prove that $(\mathbb{Z}[x],+) \cong (\mathbb{Q}^{\times},\cdot)$.
The right-hand group should be $(\mathbb Q^{\color{red}+}\color{black},\cdot)$ – that is, positive rationals, not nonzero rationals.

7. ## Interesting

Yeah, as stated in my sheet of group theory questions it is $\mathbb{Q}^{\times}$.

It must be incorrect as stated because I am pretty sure $(\mathbb{Q}^{\times}, \cdot )\not \cong (\mathbb{Q}^{+},\cdot)$.

The first has two elements of finite order $\pm 1$ while the second has only one of finite order 1.

And we have demonstrated the isomorphism to the latter.

8. Originally Posted by Gamma
It must be incorrect as stated because I am pretty sure $(\mathbb{Q}^{\times}, \cdot )\not \cong (\mathbb{Q}^{+},\cdot)$.
No, the two groups are not isomorphic. To see this, let $f:\mathbb Q^\times\to\mathbb Q^+$ be a homomorphism. Then obviously $f(1)=1$ and $1=f(1)=f((-1)(-1))=f(-1)f(-1)=f(-1)^2.$ Since $f(-1)$ must be in $\mathbb Q^+,$ we must have $f(-1)=1.$ Hence $f$ is not injective and so cannot be an isomorphism.

However $\mathbb Q^\times/\{1,-1\}\cong\mathbb Q^+$ as the mapping $f:\mathbb Q^\times\to\mathbb Q^+$ where $f(a)=a$ if $a>0$ and $f(a)=-a$ if $a<0$ is a surjective homomorphism with kernel $\{1,-1\}.$

9. Originally Posted by TheAbstractionist
By the way, just realized of something.

The right-hand group should be $(\mathbb Q^{\color{red}+}\color{black},\cdot)$ – that is, positive rationals, not nonzero rationals.
Originally Posted by Gamma
Yeah, as stated in my sheet of group theory questions it is $\mathbb{Q}^{\times}$.

It must be incorrect as stated because I am pretty sure $(\mathbb{Q}^{\times}, \cdot )\not \cong (\mathbb{Q}^{+},\cdot)$.

The first has two elements of finite order $\pm 1$ while the second has only one of finite order 1.

And we have demonstrated the isomorphism to the latter.
Originally Posted by TheAbstractionist
No, the two groups are not isomorphic. To see this, let $f:\mathbb Q^\times\to\mathbb Q^+$ be a homomorphism. Then obviously $f(1)=1$ and $1=f(1)=f((-1)(-1))=f(-1)f(-1)=f(-1)^2.$ Since $f(-1)$ must be in $\mathbb Q^+,$ we must have $f(-1)=1.$ Hence $f$ is not injective and so cannot be an isomorphism.

However $\mathbb Q^\times/\{1,-1\}\cong\mathbb Q^+$ as the mapping $f:\mathbb Q^\times\to\mathbb Q^+$ where $f(a)=a$ if $a>0$ and $f(a)=-a$ if $a<0$ is a surjective homomorphism with kernel $\{1,-1\}.$
Yes I agree, when I typed up the solution I was thinking to myself that they numbers are assumed to be positive rationals.