# Frattini subgroup of an abelian group equals to Prime subgroup

The prime subgroup $\wp(G)$ of an abelian group (maybe infinite) $G$ is defined as $\wp(G)=\bigcap_{p \in \Pi}pG$ where $\Pi$ is the set of all primes. Is it true that $\wp(G) = \Phi(G)$. Where $\Phi(G)$ is the Frattini subgroup of $G$?

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Let $G$ be an abelian group. We assume nothing about cardinality, rank, torsion etc.

Lemma: For any prime $p$, the quotient $G/pG$ is elementary abelian.

Proof. Suppose $x+pG\in G/pG$. Then $px\in pG$ so $p(x+pG)=pG$ hence $G/pG$ has exponent $p$; it is a direct sum of cyclic groups $Z_p$ (by Prüfer, Baer IIRC).

Lemma. A direct sum $\bigoplus\limits_{i\in I}C_i$ of prime-order cyclic groups has trivial Frattini subgroup.

Proof. For each coordinate $i$, let $D_i$ be the subgroup of all elements with $0$ in the $i$th coordinate; this subgroup has prime index and so is maximal; the intersection of all these maximal subgroups would have $0$ in every coordinate then, in which case it is the trivial subgroup.

Corollary. $\Phi(G)\le \wp(G)$.

Proof. By the previous two lemmas, we know that for every prime $p$, $\Phi(G/pG)=0$, hence by the lattice theorem we may lift the maximal subgroups of $G/pG$ up to maximal subgroups $M$ of $G$ such that $\bigcap M=pG$. Since the intersection of maximal subgroups is contained in $pG$ for every $p$, the corollary follows by intersecting these containments.

Lemma. $\wp(G)\le \Phi(G)$.

Proof. Suppose contrariwise $g\in\bigcap_ppG$ but $g\not\in M$ for some maximal subgroup $M<G$. Now, if the quotient $G/M$ didn't have prime order, then it would have a nontrivial proper subgroup which would lift (by lattice isomorphism theorem) to a proper subgroup of $G$ strictly containing $M$, an impossibility, so $G/M\cong Z_q$ is cyclic for some prime $q$, and $qg\in M$. Furthermore, $\langle M,g\rangle=G$.

Since $g\in\bigcap_ppG$, there is an $h\in G$ such that $g=qh$. Since $\langle M,g\rangle=G$, we have $h=m+kg$ for some $m\in M$ and $0<k<q$. But then $g=qh=q(m+kg)=qm+k(qg)\in M$ since $qg\in M$, a contradiction. Hence every element of the intersection is in every maximal subgroup.

Theorem. $\Phi(G)=\wp(G)$.

Proof. Combine the two previous propositions.

Remark. This argument is adapted from The Frattini subgroup of abelian groups.

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@azimut: Your rather zelous editing is starting to flood the front page of the group theory tag. Could you perhaps stop and continue another day? I count $7$ of the top $15$ are your edits... – user1729 Mar 27 '13 at 21:46
@user1729: ok, sorry. – azimut Mar 27 '13 at 22:00