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This is a question from my group theory exam which I was unable to prove:

It say that Show that $\mathbb Q \times C_2$ $\cong$ $\mathbb Q^*$ by specifying an isomorphism.

But I couldn't find one. How to find such a map?

(Here $\mathbb Q$ is considered a group under addition and $\mathbb Q^*$ is considered a group under multiplication.)

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    $\begingroup$ Could it be that you meant the reals (i.e., $\;\Bbb R\;$ ) instead of the rationals ($\;\Bbb Q\;$ ) ? $\endgroup$ – Timbuc Sep 8 '14 at 13:57
  • $\begingroup$ @Timbuc It is $\mathbb Q$. $\endgroup$ – spectraa Sep 8 '14 at 14:06
  • $\begingroup$ $\Bbb R^\times\cong \Bbb R\times\{\pm1\}$ via $x\mapsto(\ln|x|,x/|x)$. But $\Bbb Q\times C_2\not\cong\Bbb Q^\times$ because one side is divisible mod torsion while the other isn't. $\endgroup$ – whacka Sep 11 '14 at 4:37
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There is only one element of order $2$ in $\mathbb{Q}^*$, namely $-1$. The only element of order $2$ in $\mathbb{Q}\times C_2$ is $(0,1)$ (I'll write also $C_2=\{0,1\}$ additively). So, if $f\colon\mathbb{Q}\times C_2\to\mathbb{Q}^*$ is an isomorphism, the subgroup $H=\{(0,0),(0,1)\}$ of the domain is mapped onto $K=\{1,-1\}$, which means that $f$ induces an isomorphism $$ g\colon\frac{\mathbb{Q}\times C_2}{H}\to\frac{\mathbb{Q}^*}{K} $$ However, $\mathbb{Q}\times C_2/H\cong\mathbb{Q}$ is divisible, while $\mathbb{Q}^*/K$ is isomorphic to the group of positive rationals under multiplication which is not divisible, because it doesn't contain $\sqrt{2}$.

This is a contradiction, so $f$ doesn't exist in the first place.


What's a divisible group? An additive group $G,+$ is divisible when, for all $n>0$ and for all $g\in G$, there exists $h\in G$ such that $nh=g$. If the group is multiplicative the condition reads $h^n=g$ (just due to the different notation). Examples of divisible group are $\mathbb{Q}$, $\mathbb{R}_{>0}$ (the multiplicative group of positive reals) and $\mathbb{C}^*$ (the multiplicative group of nonzero complex numbers). Every quotient of a divisible group is divisible.

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  • $\begingroup$ what do we mean by a group being divisible? I'm new to this notion. $\endgroup$ – spectraa Sep 8 '14 at 16:02
  • $\begingroup$ @WantTobeAbstract I added the definition and some examples. $\endgroup$ – egreg Sep 8 '14 at 16:06
  • $\begingroup$ I was too quick answering to the post and deleted my wrong argument. Egreg's is fine! $\endgroup$ – Nicky Hekster Sep 8 '14 at 17:53
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You are to be congratulated for not finding an isomorphism, as these groups are not isomorphic. The multiplicative group $\mathbb{Q}^{\ast}$ is isomorphic to $C_2\times\mathbb{Z}^{\omega}$, a direct sum of cyclic groups, while the additive group $\mathbb{Q}$ is divisible.

Added: If, according to @Timbuc's suggestion, it was $\mathbb{R}$ that was meant, then there is indeed an isomorphism. If we think of $C_2$ as the group $\{1,-1\}$ under multiplication, then an isomorphism $\mathbb{R}\times C_2 \to \mathbb{R}^{\ast}$ is given by $(r,\sigma)\mapsto \sigma e^r$.

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  • $\begingroup$ but the question paper specified so certainly that - show that $\mathbb Q$ X $C_2$ $\cong$ $\mathbb Q^*$ and specifying the reason why the function you give is indeed an isomorphism. $\endgroup$ – spectraa Sep 8 '14 at 13:50
  • $\begingroup$ by the way I found this new group $\mathbb{Z}^{\omega}$ .What is this group?can you please explain. $\endgroup$ – spectraa Sep 8 '14 at 13:52
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    $\begingroup$ This group $\mathbb{Z}^\omega$ is just a direct sum of infinitely many copies of $\mathbb{Z}$; i.e., a free abelian group of countably infinite rank. $\endgroup$ – James Sep 8 '14 at 14:01

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