Prove that $\displaystyle (\mathbb{Z}[x],+) \cong (\mathbb{Q}^{\times},\cdot)$.
I am not seeing it.
If $\displaystyle a\in \mathbb{Q}^{\times}$ we can write $\displaystyle a = \prod_{n=0}^{\infty} p_n^{e_n}$ where $\displaystyle e_n\in \mathbb{Z}$ and $\displaystyle p_1,p_2,p_3,...,$ are the primes in increasing order.
Thus, I think the natural map is $\displaystyle \phi \left( \sum_{n=0}^{\infty} a_n x^n \right) = \prod_{n=0}^{\infty} p_n^{e_n}$.
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 $\displaystyle \phi \left( \sum_{n=0}^{\infty}a_n x^n + \sum_{n=0}^{\infty}b_n x^n \right) = \phi \left( \sum_{n=0}^{\infty} a_n x^n \right) \phi \left( \sum_{n=0}^{\infty} b_n x^n \right)$ since addition of exponents is adding the exponents.
This happens to be one-to-one and onto.
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 $\displaystyle 2^{a_1}3^{a_2}5^{a_3}\cdots$ where $\displaystyle a_i\in\mathbb Z$ is very beautiful indeed.
Yeah, as stated in my sheet of group theory questions it is $\displaystyle \mathbb{Q}^{\times}$.
It must be incorrect as stated because I am pretty sure $\displaystyle (\mathbb{Q}^{\times}, \cdot )\not \cong (\mathbb{Q}^{+},\cdot)$.
The first has two elements of finite order $\displaystyle \pm 1$ while the second has only one of finite order 1.
And we have demonstrated the isomorphism to the latter.
No, the two groups are not isomorphic. To see this, let $\displaystyle f:\mathbb Q^\times\to\mathbb Q^+$ be a homomorphism. Then obviously $\displaystyle f(1)=1$ and $\displaystyle 1=f(1)=f((-1)(-1))=f(-1)f(-1)=f(-1)^2.$ Since $\displaystyle f(-1)$ must be in $\displaystyle \mathbb Q^+,$ we must have $\displaystyle f(-1)=1.$ Hence $\displaystyle f$ is not injective and so cannot be an isomorphism.
However $\displaystyle \mathbb Q^\times/\{1,-1\}\cong\mathbb Q^+$ as the mapping $\displaystyle f:\mathbb Q^\times\to\mathbb Q^+$ where $\displaystyle f(a)=a$ if $\displaystyle a>0$ and $\displaystyle f(a)=-a$ if $\displaystyle a<0$ is a surjective homomorphism with kernel $\displaystyle \{1,-1\}.$