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I've already proved, that there are no non-trivial subgroups with a trivial intersection.

E.g. $H,G$ are subgroups in $F$, $\dfrac{a}{b} \in H$ and $\dfrac{c}{d} \in G$, $\dfrac{a}{b}+\dfrac{a}{b}+...+\dfrac{a}{b}=\dfrac{ab}{b} \in H,a \in H$ $a+a+...+a=ac \in H$. Same for $c \in G$

I guess, I need to show, that if $F \simeq G \times H$ ,then there are $H_{1} \le H, G_{1} \le G: H_{1} \cap G_{1} = \{e\} $ and $H_{1} \simeq H,G_{1} \simeq G$

No idea how to do it.

A-grin
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1 Answers1

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The statement should be that $\mathbb{Q}$ is not the direct product of two proper subgroups.

The idea is correct. If you have two nonzero subgroups $H_1$ and $H_2$ and take nonzero $x=a/b\in H_1$, $y=c/d\in H_2$, then you have $$ z=bcx=ady\in H_1\cap H_2 $$ because subgroups are closed under multiples and clearly $z\ne0$.

Therefore no two nonzero subgroups intersect trivially and you're done.

egreg
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  • How do I prove, that such subgroups with mentioned features exist though? – A-grin Jun 27 '21 at 20:02
  • @A-grin Do you need to? If $\mathbb{Q}$ is the direct product of $K_1$ and $K_2$, then $K_1\cap K_2={0}$ by definition, so by what we proved, one of them is the trivial subgroup. – egreg Jun 27 '21 at 20:17
  • Unfortunatelly, there is no definition for the direct product of subgroups in my course, only for the direct product for groups and as we know, groups can intersect non-trivially in the last case. So I can't use it. – A-grin Jun 27 '21 at 20:25