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Frattini subgroup $Φ ( G )$ of a group G is the intersection of all maximal subgroups of $G$. The question is

Let $G$ be an abelian group.What is the sufficient and necessary condition of $Φ ( G ) ≠｛e｝$？

I think if $G$ has no torsion part, we cannot say $Φ ( G ) ≠｛e｝$,because $\Bbb Z ×\Bbb Z×・・・ ×\Bbb Z$ has a maximal subgroup $p\Bbb Z ×\Bbb Z×・・・×\Bbb Z$, with intersectioning distinct primes, we gain just a trivial group.

So, we need torsion part. But I cannot proceed from here.I only know the answer, $G$ has an element of order $p^2$, that is, $G$ has theree subgroups. I want to deduce this consequence from steady discussion.

P.S The Frattini subgroup of $\Bbb{Z}_p \times\Bbb Z _{p^2}.$ According to this answer, in a ring that is a finite direct product of quotient rings of $\mathbf{Z}$, we can say rattini subgroup is the same as the Jacobson radical of the corresponding ring.

Poitou-Tate
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  • Use $\Phi$ for $\Phi$, $\times$ for $\times$, and $\neq$ for $\neq$. – Shaun Nov 10 '20 at 16:54
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    Are you aware of the fact that the Frattini subgroup consists of the "non-generators" of the group? (Elements which never occur in a minimal generating set of the group.) So prove that if $x$ has order $p^2$ then $x^p$ is a non-generator (which is manageable). – user1729 Nov 10 '20 at 22:12
  • Note that if $A$ and $B$ are groups, then $\Phi(A\times B)\subseteq \Phi(A)\times\Phi(B)$: if $M$ is maximal in $A$, then $M\times B$ is maximal in $A\times B$, and similarly for $B$. Similar for a finite product. It should be easy to figure out $\Phi(\mathbb{Z}$ and $\Phi(\mathbb{Z}_{p^a})$ for primes $p$. That will give you an upper bound. – Arturo Magidin Nov 11 '20 at 02:44
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    Why is this a different question from this one ? – Arturo Magidin Nov 11 '20 at 03:27
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    As I said in a comment to your earlier post, we seem to be just talking about finitely generated abelian groups. I don't know what the answer is for abelian groups that are not finitely generated. – Derek Holt Nov 11 '20 at 07:58

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