# Frattini subgroup of additive group of rational numbers

Show that Frattini subgroup of additive group of rational numbers $(\mathbb{Q},+)$ is itself, or $$\Phi(\mathbb{Q})=\mathbb{Q}$$

PS. My strategy is prove that group $(\mathbb{Q},+)$ hasn't maximal normal subgroup. Assume that $\mathbb{Q}$ has maximal normal subgroup $M$, then since $(\mathbb{Q},+)$ abel, so nilpotent, and satisfies the normalier condition. thus $M$ is a normal subgroup of $\mathbb{Q}$ and has index of prime.

Can you help me show that $\mathbb{Q}$ hasn't a subgroup with index of prime? Thanks.

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Are you sure that the normalizer condition holds for infinite nilpotent groups? – Ted Oct 17 '12 at 8:57
Yes, if $G$ is nilpotent, then it satisfies the normalizer condition – Muniain Oct 17 '12 at 9:12
Why do you want to use nilpitency when you have an abelian group? Any subgroup is normal, so just show that the rationals do not have any proper maximal subgroup. – Tobias Kildetoft Oct 17 '12 at 9:26
nilpitency imply $M$ has index of prime – Muniain Oct 17 '12 at 10:13
@Muniain So does abelian, and it is much easier to see that this is the case. – Tobias Kildetoft Oct 17 '12 at 11:27

## 3 Answers

Hints:

1) $\,\Bbb Q\,$ is a divisible group, meaning:

$$\forall\,q\in\Bbb Q\,\,\,and\,\,\,\forall\,0<n\in\Bbb N\,\,\,\exists\,r\in\Bbb Q\,\,\,s.t.\,\,\,q=nr$$

2) A homomorphic image of a divisible group is divisible

3) A finite group cannot be divisible

4) A normal subgroup $\,M\,$ of a group $\,G\,$ is a maximal proper subgroup iff $\,G/M\,$ is finite of order a prime

From all the above, it then follows that $\,\Bbb Q\,$ cannot have maximal subgroups.

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Let $G$ be a nontrivial proper subgroup of $\Bbb Q$ (it is automatically normal, since $\Bbb Q$ is commutative). We show that there is a $H\supsetneq G$ proper subgroup.

First, $G$ contains a nonzero integer $a$ (since $G$ has some nonzero elemet $a/b$, so $a\in G$ and $-a\in G$ as well). There is a smallest positive among them, we can assume that $a$ is such. If $a> 1$, then $G/a:=\{\frac xa\mid x\in G\}$ will be good: it still doesn't contain $1/a$, but already contains $1$.

Hence, we can assume that $a=1$, i.e. $\Bbb Z\subseteq G$. Then let $b:=\min\{n>0\mid \frac1n\notin G\}$. Then $G/b$ will be a good intermediate subgroup again.

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How can we exclude that your definition of $b$ leads to $\infty$, i.e. show that $\{n>0|\frac1n\not\in G\}\neq\varnothing$? Does the set of unit fractions generate the rationals? – MickG Aug 3 '15 at 13:34

For abelian groups, any (proper) maximal subgroup has finite index (in fact prime index), for if $M$ is a maximal subgroup of $G$, and (as subgroups of abelian group are normal) if $G/M$ is infinite, then it contains a proper non-trivial subgroup, contradicting maximality of $M$. Therefore, it is sufficient to show that $\mathbb{Q}$ has no subgroups of finite index. For, if $M$ is subgroup of index $k\geq 1$ in $\mathbb{Q}$, consider any rational $a/b$ in reduced form. As $\mathbb{Q}/M$ has order $k$, every element of $\mathbb{Q}$ added $k$ times to $0$ will fall in $M$. But, $a/b=k(a/kb)\in M$, i.e. $\mathbb{Q}=M$. q.e.d

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