9
$\begingroup$

$\mathbb{Q}^{\times}$ is the group of rational number without $0$ under multiplication, and $\mathbb{Z}^{n}$ is the free abelian group of rank $n$. Show that $\mathbb{Q}^{\times}$ not isomorphic to $\mathbb{Z}^{n}$. I tried to assume that there is an isomorphism and to get a contradiction.. didn't succeed

$\endgroup$
  • 8
    $\begingroup$ $\Bbb Q^\times$ has torsion, but free abelian groups do not. In fact $\Bbb Q^\times$ is isomorphic to $\Bbb Z_2\oplus \bigoplus \Bbb Z$ where the sum is over countably many copies of the integers. $\endgroup$ – Pedro Tamaroff Jan 18 '15 at 16:47
  • $\begingroup$ That's even better, you should put this as an answer, even if it same the same as @Meelo wrote, and as what anyone could wrote. (Bet.) $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 18 '15 at 18:17
11
$\begingroup$

Suppose you had a homomorphism $f:\mathbb Q^{\times}\rightarrow\mathbb Z^{n}$. Notice that $(-1)^2=1$ thus $f(-1)^2=1$. The only solution to $x^2=e$ (where $e$ is the identity) in $\mathbb Z^{n}$ is $e$, thus $f(-1)=f(1)=e$, meaning no homomorphism is injective and the groups are not isomorphic.

(To be clear, notice that $x^2=e$ in $\mathbb Z^n$ would more commonly be written as $x+x=0$, but I chose to write it multiplicatively to aid seeing the connection between $(-1)^2=1$ (in $\mathbb Q^{\times}$) and $x^2=e$ (in $\mathbb Z^{n}$))

$\endgroup$
  • $\begingroup$ It is very unorthodox to denote the zero of $\Bbb Z^n$ as $1$. =) $\endgroup$ – Pedro Tamaroff Jan 18 '15 at 16:50
  • $\begingroup$ I edited its answer properly to address that, but the guy apparently refused my edit... $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 18 '15 at 16:52
  • $\begingroup$ @RobertGreen It is better to comment to the poster on such minor detail (which is not incorrect!) rather than to "force" the edit oneself. $\endgroup$ – Pedro Tamaroff Jan 18 '15 at 16:53
  • $\begingroup$ I would tend to see this as coding with other people. The code does not belong to anyone, and anyone is free to "edit" change it, provided that this edit preserves the function/results etc of the code. Btw, I'm not forcing anything, as he can (and he did) refuse. What I say is that my edit is valid as it corrects what he put in the first place. Instead of validating the edit, he preferred to give an answer that is stil really cumbersome : "$x^2 = e$ in $\mathbf{Z}^n$"... In which serious place have you ever seen this kind of stuff ? ;-) $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 18 '15 at 16:56
  • 1
    $\begingroup$ @Robert It had not initially occurred to me that the notation $x^2=e$ was at all unnatural and, after seeing your edit, I chose to keep the notation, but to note that it was not standard. I didn't like your edit because it didn't seem helpful to say one group is multiplicative and the other is additive - those properties aren't intrinsic to the group, so I'm unconcerned with them (esp. since I usually think of $\mathbb Z^n\cong \langle a_1,\ldots,a_n | a_ia_j=a_ja_i\rangle$). Your edit was valid, but it conflicted with my intention - a comment (or even separate answer) would have been better. $\endgroup$ – Milo Brandt Jan 18 '15 at 17:41
11
$\begingroup$

$\mathbf{Q}^{\times}$ is generated by the set of positive prime numbers and $-1$ so it contains (a groupe isomorphic to) $\mathbf{Z}^{\mathscr{P}}$ where $\mathscr{P}$ is the set of positive prime numbers. $\mathbf{Z}^{\mathscr{P}}$ is not of finite rank over $\mathbf{Z}$ whereas any $\mathbf{Z}^n$ is, so there's no possible isomorphism $\mathbf{Q}^{\times} \rightarrow \mathbf{Z}^n$, for any $n$, as any such arrow would give an infinite free family in $\mathbf{Z}^n$, namely, the family image of $\mathscr{P}$ by the supposed isomorphism.

To put if in a "fashion" way the morphism $$v : \mathbf{Q}^{\times} \rightarrow \{\pm\}\times \mathbf{Z}^{\mathscr{P}}$$ defined by $v(x) = \left( \textrm{sign}(x), \left( v_p(x)\right)_{p\in \mathscr{P}} \right)$ where $v_p$ is the $p$-adic valuation function for $p\in \mathscr{P}$, is an isomorphism.

This relies on the fact that $\mathscr{P}$ is infinite - fact admitted here. ;-)

$\endgroup$
  • 1
    $\begingroup$ Your argument is that $\mathbf Q^\times_{>0}$ not finitely generated while $\mathbf Z^n$ is. Although this is correct, showing that there is no set of generators of cardinality less than the cardinality of the set of primes (and $-1$) can be unpleasant. $\endgroup$ – Myself Jan 18 '15 at 16:40
  • $\begingroup$ @Myself Why? It's the fundamental theorem in action; clearly if a given $p_i$ appears in a member of $\mathbb{Q}^\times$ then it has to be in one of the generators... $\endgroup$ – Steven Stadnicki Jan 18 '15 at 16:47
  • $\begingroup$ Is it really that unpleasant? Any finite set of elements of ${\mathbb Q}^\times$ can only involve finitely many primes in their numerators and denominators, so cannot generate the whole group. $\endgroup$ – Derek Holt Jan 18 '15 at 16:47
  • $\begingroup$ @Myself forgot to put $-1$, sorry. Nothing is unpleasant actually, as you may see in my edited answer $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 18 '15 at 16:48
  • $\begingroup$ Ah maybe it's not quite so unpleasant as I thought :-) For a minute I thought there would be lcm's and gcd's all over the place; but that is for $(\Bbb Q,+)$. $\endgroup$ – Myself Jan 18 '15 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.