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Can anyone please help me to find the Frattini subgroup of $\mathbb{Z}_p \times \Bbb Z _{p^2}$? I know that as a set the Frattini subgroup is the set of all non-generators. Is this the only way to compute such subgroups? Is there any better way? My professor's answer was negative.

I believe there should be some cool way to tackle this problem. Can we solve this problem from the definition? I mean by listing all the maximal subgroups, and taking their intersection?

Thanks so much.

Shaun
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Gordhob Brain
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2 Answers2

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The Frattini subgroup of $\mathbf{Z}_p \times \mathbf{Z}_{p^2}$ is equivalently its Jacobson radical as a ring (this is actually true for any ring that is a finite direct product of quotient rings of $\mathbf{Z}$). Since it is a finite ring, the Jacobson radical is the same as the nilradical, which is $\{0\}\times p\mathbf{Z}_{p^2}$. Hence, the Frattini subgroup of $\mathbf{Z}_p \times \mathbf{Z}_{p^2}$ is also $\{0\}\times p\mathbf{Z}_{p^2}$.

Geoffrey Trang
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One way to solve this problem is to know that the Frattini subgroup behaves well with respect to direct products. That is, for finite groups $A$ and $B$ we have $\Phi(A \times B) = \Phi(A) \times \Phi(B)$. Applying this with $A=C_p$ and $B=C_{p^2}$ yields $\Phi(C_p \times C_{p^2}) = K$, where $K$ is the unique subgroup of order $p$ of $B$.

Working with the definition: note that $\Phi(C_p \times C_{p^2})>1$ and that $C_p \times K$ and $1 \times C_{p^2}$ are both maximal subgroups, thus $\Phi(C_p \times C_{p^2}) \leq K = \left(C_p \times K\right) \cap \left(1 \times C_{p^2}\right)$. It follows that $\Phi(C_p \times C_{p^2}) = K$.

the_fox
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  • This $K$ is isomorphic to $C_p$, right? This simply Cauchy theorem. Am I correct? Thanks so much for your answer. – Gordhob Brain Feb 04 '20 at 05:16
  • In mathoverflow, I have seen that $\Phi(A \times B) = \Phi(A) \times \Phi(B)$ is not always true Please have a look. https://mathoverflow.net/questions/283989/product-of-frattini-groups – Gordhob Brain Feb 04 '20 at 05:27
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    It is definitely true for finite groups and I think even for finitely generated groups, but not for arbitrary groups. This is a theorem of Gaschütz, see here for instance: https://math.stackexchange.com/questions/1026149/translation-of-the-statements-of-theorems-presented-in-a-german-paper-of-gaschüt – the_fox Feb 04 '20 at 05:34
  • Could you explain it a bit? How do you know that $\Phi(C_p \times C_{p^2})>1$? Thanks again. Also, we need to show that these two are the only maximal subgroups. How can we argue that? – Gordhob Brain Feb 04 '20 at 05:38
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    It is a standard theorem that a (finite) $p$-group $G$ has trivial Frattini subgroup if and only if it is elementary abelian. One way to see this is to observe that $G^p = \langle g^p: g \in G \rangle$ is contained in $\Phi(G)$. This holds because a maximal subgroup of a finite $p$-group has index $p$ in the group so if $M$ is maximal in $G$ then $g^pM = (gM)^p = M$, so $g^p \in M$ for all $g \in G$. – the_fox Feb 04 '20 at 05:42
  • No, those two are not the only maximal subgroups. In fact, $C_p \times C_{p^2}$ has $p+1$ maximal subgroups. But the intersection of just two of them contains the intersection of all of them. – the_fox Feb 04 '20 at 05:47