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The group $\mathbb Q ^{*}$ is the group of all rational numbers under the multiplication operation. $N = \{-1,1\}$ is a normal subgroup of $\mathbb Q ^{*}$. $\mathbb Q ^{+}$ is a subgroup of $\mathbb Q ^{*}$, where $\mathbb Q ^{+}$ is the group of all positive rational numbers.

How would I use the first isomorphism theorem to show that $\mathbb Q ^{*} / N$ is isomorphic to $\mathbb Q ^{+}$?

I was going to start off by creating a surjective homomorphism map from $\mathbb Q ^{*}$ to $\mathbb Q ^{+}$ that has a kernel $N$.

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    $\begingroup$ That's all you have to do. $\endgroup$ – Couchy311 Nov 20 '16 at 4:10
  • $\begingroup$ But how would that show isomorphism? $\endgroup$ – PiccolMan Nov 20 '16 at 4:13
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    $\begingroup$ If $G\to H$ is a surjective homomorphism with kernel $K$, then the isomorphism theorem tells you that $G/K\cong H$. The only difficulty, if any, will be to define the map $\mathbb Q^*\to \mathbb Q^+$. $\endgroup$ – Couchy311 Nov 20 '16 at 4:15
  • $\begingroup$ Does K also have to be normal to G? $\endgroup$ – PiccolMan Nov 21 '16 at 2:46
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    $\begingroup$ I just realized that the kernel is always normal. $\endgroup$ – PiccolMan Nov 21 '16 at 2:49
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Let's consider the following map: $\phi: \mathbb{Q^{*}} \rightarrow \mathbb{Q^{+}}$, where $\phi(a)=|a|$. Clearly, this map is homomorphism and epimorphism also (check!). Now, if $x \in \ker\phi $, then $|x|=1$, so that $x=\pm1$ and $\ker\phi=N$. So that we are done.

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