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I have to prove that the group $(\mathbb{Q},+)$ hasn't got any maximal subgroup.

Let $H$ be a maximal subgroup of $G=\mathbb{Q}$. So $H$ is normal in $G$ and I can consider the quotient group $G/H$. My idea is the following: if I prove

1) this quotient hasn't got any maximal subgroup

2) this quotient has order $p$, where $p$ is a prime number

then I could consider $r \in G$ and obtain $pr \in H$. Therefore $p \frac{x}{p}=x \in H$ for any $x \in G$, so $H=G$, absurde!

But I can't prove 1) and 2)...

federico
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  • I have proved the first request, but I can't prove the second one, that's to say why $\mathbb{Q}/H$ has order $p$ for some prime number $p$? –  Oct 31 '14 at 12:15

2 Answers2

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  1. $G/H$ has no proper non-trivial subgroup because any such subgroup would come from a subgroup of $G$ containing $H$, and $G$ and $H$ are the only possible choices because $H$ is maximal.

  2. A non-trivial group $\Gamma$ with no proper non-trivial subgroup must be cyclic because any non-trivial element generates a subgroup. The group $\Gamma=\langle g \rangle$ cannot be infinite because $\langle g^2 \rangle$ would be a proper subgroup. Hence, $\Gamma$ is finite of order $n$. If $n=ab$ were composite, then $\langle g^a \rangle$ would be a proper subgroup. Thus, $\Gamma$ has prime order. Apply to $\Gamma=G/H$.

lhf
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If you start with any strict subgroup $H$ of the rationals, can's you always generate a new subgroup with $H$ and a rational number not in $H$?

Simon S
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