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Definition. A maximal subgroup of a group G is a proper subgroup $M$ of $G$ such that there are no subgroup $H$ of $G$ such that $M<H<G$.

Now, I want to solve the following problem:

Problem: $(\mathbb{Q},+)$ has no maximal subgroup.

There are solutions available here. But I find these hard to me...So I try to solve this own...

My Solution: Suppose that $M$ be a maximal subgroup of $(\mathbb{Q},+)$. Then by Fourth Isomorphism Theorem the subgroups of $\mathbb{Q}/M$ are only $\{0\}$ and $\mathbb{Q}/M$. And then $\mathbb{Q}/M$ must be a finite group, Since it can be proved that A group is finite if and only if it has finite number of subgroups (See here). Therefore $[\mathbb{Q}:M]=|\mathbb{Q}/M|<\infty$. Therefore $M$ becomes a proper subgroup of finite index of $(\mathbb{Q},+)$, and which, we know, is not possible. (*because, $(\mathbb{Q},+)$ has no proper subgroup of finite index). Hence we get a contradiction.

*Theorem. $(\mathbb{Q},+)$ has no proper subgroup of finite index

Proof: Suppose $H \le \mathbb{Q}$ such that $[\mathbb{Q}:H]=|\mathbb{Q}/H|=n$ (say) Now let $q \in \mathbb{Q}$. Then $(q+H)^n=H$ i.e. $nq+H=H$ i.e., $nq \in H$. Since $q$ is arbitrary in $\mathbb{Q}$ so this means $n\mathbb{Q} \subset H$. But look $n\mathbb{Q}=\mathbb{Q}$, since $q \mapsto nq$ is an automorphism of $\mathbb{Q}$. Hence $H=\mathbb{Q}$.

Is this solution correct? Thank you..

sigma
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    You're confusing the number of subgroups of $\Bbb Q/M$ with the number of elements (i.e. $|\Bbb Q/M|$). And how do you know $\Bbb Q$ has no subgroups of finite index$? – anon Apr 22 '18 at 02:43
  • Oho sry sry Thanks ...I'll edit it now.. – sigma Apr 22 '18 at 02:45
  • @anon Is everything right now..? – sigma Apr 22 '18 at 02:48
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    I know that $\Bbb Q$ has no subgroup of finite index, but can you prove that, rather than simply assert it. – Angina Seng Apr 22 '18 at 02:50
  • Yes, I can Prove it .....ok I'll edit it in my post. – sigma Apr 22 '18 at 02:51
  • @LordSharktheUnknown Does this suffice the solution of my original problem..? – sigma Apr 22 '18 at 03:08
  • If the solution given in the link is not easy enough to understand, here is a rephrased one: Suppose $H<\mathbb Q$ is maximal, then choose any rational $x\notin H$, the subgroup $H+\langle x\rangle$ should be $\mathbb Q$ by maximality. However, for any $y\in H$, $y/x=a/b$ for some integer $a,b$, and so if $x/a=h+nx$ for some $h\in H, n\in\mathbb N$, $$x=ah+n(ax)=ah+nby\in H$$, but it is false by definition of $x$. Hence $\mathbb Q\ni x/a\notin H+\langle x\rangle$, which is contrary to our assertion that $H+\langle x\rangle=\mathbb Q$, we hence arrive at a contradiction. – BAI Apr 22 '18 at 10:28
  • ($y\neq 0$ in the previous comment) – BAI Apr 22 '18 at 10:35

1 Answers1

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Your proof is good, but the first part can be simplified.

If $M$ is a maximal subgroup, the quotient $\mathbb{Q}/M$ is a simple abelian group, hence certainly finite (actually of prime order): any of its non identity element is a generator, so the group is cyclic and it cannot be infinite cyclic.

Now the task is to show the stronger statement that $\mathbb{Q}$ has no proper subgroup of finite index.

If $H$ is a subgroup of finite index $n$, then $n\mathbb{Q}\subseteq H$: indeed, for every $q\in\mathbb{Q}$, $n(q+H)=H$, which amounts to saying that $nq\in H$. However, $n\mathbb{Q}=\mathbb{Q}$, so $H=\mathbb{Q}$.

You can note that no divisible abelian group has proper subgroups of finite index, with exactly the same proof as for $\mathbb{Q}$.

egreg
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  • nice one....thank you ... – sigma Jun 16 '18 at 07:45
  • If $M$ is a maximal subgroup of $\Bbb Q$, why must $\Bbb Q/M$ necessarily be a simple abelian group? Also, why is it necessarily finite? –  May 02 '20 at 15:04
  • Actually, I think it has to be simple abelian because any subgroup of an abelian group is normal. Moreover, any quotient of an abelian group is abelian. Moreover, we're picking the maximal normal subgroup so the quotient has to be a simple abelian group. However, how do we know simple abelian groups are necessarily finite? –  May 02 '20 at 15:09
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    @Triskelion Since the subgroup is normal, the quotient group cannot have proper nontrivial subgroups, so it is simple (and abelian, because $\mathbb{Q}$ is abelian). A simple abelian group is cyclic, as explained in my answer. The only infinite cyclic group is not simple. – egreg May 02 '20 at 15:09
  • I don't understand why for every $q \in \Bbb Q$, $n(q + H) = H$. Could you please explain that part? I understand that $q + H$ gives some coset of $H$ in $\Bbb Q$. I mean, it might make sense to write $H + nq = H$ because $n$ helps to traverse all the possible cosets in $\Bbb Q$ and return back to the original subgroup coset $H$. But how do we know $nH = H$? –  May 02 '20 at 15:35
  • @Triskelion Note that $H=0+H$ is the zero element in $\mathbb{Q}/H$. This is nothing else than $nx=0$ in an abstract additive group. – egreg May 02 '20 at 15:38
  • Ah, thanks. I guess that explains why $H + nq = H$ (because the cosets are $H + 0q, H+q, H+2q, \ldots, H + (n-1)q$). But I'm still not sure why $nH = H$...also why $n\Bbb Q = \Bbb Q$? –  May 02 '20 at 15:41
  • @Triskelion From $\mathbb{Q}=n\mathbb{Q}\subseteq nH$ we get $\mathbb{Q}\subseteq nH\subseteq H$. – egreg May 02 '20 at 16:04