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I have a problem that I don't have any idea.

show that group $(\mathbb{Q},+)$ has no maximal subgroup.

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What's a maximal subgroup? – Matt N. Nov 11 '12 at 17:06
No maximal proper subgroup, I think. What does a subgroup look like? – Mark Bennet Nov 11 '12 at 17:08
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I suggest to work on your accept rate – Belgi Nov 11 '12 at 17:09
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definition maximal subgroup is proper subgroup – Firmino Nov 12 '12 at 1:25

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up vote 7 down vote accepted

Suppose $H$ is any nonzero proper subgroup of $\mathbb Q$ and let $x \in \mathbb Q \setminus H$ and $y \in H, y \neq 0$

Write $y/x = a/b$ with integers $a,b$. Then $a \neq 0$ and $x/a \notin H + \langle x \rangle$ : Suppose $x/a = h + nx$ for some $n \in \mathbb Z$ and $h \in H$. Then $x = ah+anx = ah+nby \in H$, which contradicts the hypothesis on $x$. Thus $H$ is not maximal.

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Are you assume that $H$ is maximal subgroup of $\mathbb{Q}$? – Firmino Nov 12 '12 at 9:13
@Firmino : no I pick any proper subgroup and show that it is not maximal. – mercio Nov 12 '12 at 9:27
and use $H+\langle x \rangle = \mathbb{Q}$ to implies $x/a \notin \mathbb{Q}$, contradicts?. Thank you so much. – Firmino Nov 12 '12 at 16:18
@Firmino : Since $x/a \notin H+\langle x \rangle = H'$, $H'$ is a proper subgroup of $\mathbb Q$, and since $x \in H'$ and $x \notin H$, $H$ is strictly smaller than $H'$, thus $H$ isn't a maximal proper subgroup. – mercio Nov 12 '12 at 19:06
I think $H+\langle x \rangle =\mathbb{Q}$. If reverse then we have $H+\langle x \rangle$ is proper subgroup of $\mathbb{Q}$, implies $H$ is proper subgroup of $\mathbb{Q} - \langle x \rangle$. contradict with maximal property of $H$. – Firmino Nov 13 '12 at 11:05
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up vote 12 down vote

Hint: Assume by contradiction that $H$ is a maximal subgroup of $\mathbb Q$. Pick some $x \in \mathbb Q \backslash H$.

Now, since $H$ is maximal then $H + \langle x \rangle = \mathbb Q$.

All you have to do now is look at $\frac{x}{2}$.

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Are you suggesting that for any subgroup $H$ and any element $x \notin H$, then $x/2 \notin H+<x>$ ? because that's not true. – mercio Nov 11 '12 at 17:19
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+1. No contradiction is required btw. Your argument shows that any subgroup that is not $\mathbb{Q}$ is contained in a larger subgroup that is also not $\mathbb{Q}$ and can therefore not be maximal. – WimC Nov 11 '12 at 17:20
@mercio - could you give an example? – Mark Bennet Nov 11 '12 at 17:39
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Well, pick $H = \langle \frac 1 2 \rangle$ and $x = \frac 1 3$. I'm pretty sure $\frac 1 6 = \frac 1 2 - \frac 1 3$ – mercio Nov 11 '12 at 17:40
@mercio - that makes your comment a lot clearer, thanks – Mark Bennet Nov 11 '12 at 17:41
up vote 1 down vote

The group $\mathbb Q$ is a divisible group. There is a well known facts that says $G$ is divisible if and only if $G$ has no maximal subgroups if and only if every nonzero quotient of $G$ is infinite.

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Very simple, sound, explanation +1 – amWhy Feb 6 at 0:09
@amWhy: Thanks for kind words. – Babak S. Feb 6 at 2:20

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