# $(\mathbb{Q},+)$ has no maximal subgroup

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
I suggest to work on your accept rate –  Belgi Nov 11 '12 at 17:09
definition maximal subgroup is proper subgroup –  Firmino Nov 12 '12 at 1:25

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

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
+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
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
@N.S. Could you clarify a little more? –  Vishal Jun 13 at 6:54
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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