Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 5 down vote accepted


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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.