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Show that the group of rational numbers $\Bbb Q$ doesn't contain non-trivial subgroups $G$ and $H$ such that $\Bbb Q$ is isomorphic to $G \oplus H$

Let $R$ be $\Bbb Z_2[x]/\langle x^2 + 1 \rangle$. Prove that $R$ is not isomorphic to $\Bbb Z_4$.

I think I need some help in doing not isomorphism. How to prove that stuff?

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    $\begingroup$ You ask two different questions. It is better that you ask two questions. $\endgroup$ – Babak Miraftab Jan 20 '14 at 19:46
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Hint: If $r$ is a rational number then there exists some non-zero integer in the subgroup generated by $r$.

Thus, each of $G,H$ must contain an integer. What can you say about a common multiple of those two integers?

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$\textbf{Hint:}$ For the second part of your query, you could write out the elements of $R$ explicitly. $R=\{a_0+a_1x : a_i \in \mathbb{Z}_2\}$, since we have $x^2+1=0 \Rightarrow x^2 \equiv 1$. This gives just four elements in $R$. Now check what kind of relations exist between these elements.

For example, how many generators does $R$ need? Is $R$ cyclic? Is $\mathbb{Z}_4$ cyclic?

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A possible hint for your first question. We know that for every group $G$, there is a decomposition like $$G=dG\oplus R$$ in which $R$ is reduced. Moreover $dG$ is divisible and so $dG=G$. This makes $R$ to be trivial. I hope this helps.

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  • $\begingroup$ Dear Babak, $dG$ is a subgroup when $G$ is an abelian group. Furthermore, we can only decompose every abelian group to $G=dG\oplus R$ $\endgroup$ – Babak Miraftab Jan 21 '14 at 7:14

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