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I would like to ask two questions. Kindly help me in this regard:

(1) In Finite p-group with a cyclic frattini subgroup., user28083 has given finite $p$-group with cyclic frattini subgroup. I need the reference of this.

(2) Suppose $G$ is group of order either $p^4$ or $p^5$ ($p$ a prime) such that all subgroups of order $p$ of non cyclic frattini subgroup are normal in $G$. What are the choice for $G$? I don't want to use the classification of these. Could you kindly provide a simple argument for that?

with regards

Vipul

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You mean you need a reference for the parts in that question given in the images? For your second question, you mean that $G$ has a non-cyclic Frattini subgroup, and any subgroup of order $p$ contained in it is normal? –  Tobias Kildetoft Aug 20 '13 at 8:39
    
For @Tobias' re-writing of the problem, you should look up the Burnside Basis Theorem. This states that if $|G: \Phi(G)|=p^r$ then every set of generators of $G$ has a subset of $r$ generators which also generates $G$. So for $p^4$, non-cyclic Frattini implies that $G$ is precisely two generated and so $\Phi(G)$ has order $p^2$ and so is cyclic. So $p^4$ cannot happen. For $p=5$, you need two or three generated, and then three-generated gives you the $p^2$ contradiction. So $G$ is necessarily two-generated. I wonder if the normality bit is superfluous...it is for $p=2$... –  user1729 Aug 20 '13 at 8:59
    
Dear user1729. I am not able to understand how two generator would imply that $\Phi(G)$ is cylcic for $p^4$. –  Vipul Kakkar Aug 20 '13 at 9:22
    
Sorry, that is a mistake. If $G$ is two-generated then $\Phi(G)$ is abelian (not cyclic), because every group of order $p^2$ is abelian (as $G/Z(G)$ and $Z(G)$ are both cyclic, which implies that $Z(G)=G$). (Also, if you put @ before someone's name then they will be pinged. You are pinged automatically here as this is your question, which is why I have not been doing it with you.) –  user1729 Aug 20 '13 at 9:43
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