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Theorem. Let $G$ be a finite, non-abelian $p$-group all of whose proper subgroups are abelian. Then $|G'|=p$.

Take a counterexample of minimal order. Assume that exist a $H$ such that $1 Then (by $G'\leq \Phi (G) \leq Z(G)$) $H\vartriangleleft G$. From this we deduce we can assume $|G'|\leq p^2$.

Then? How am I supposed to continue?

Edit
Additional infos
$G'$ is elementary abelian since $G$ is Frattini-in-center.

  • 1
    We see this nice deduction when $|G|=p^3$ and the group is non-abelian and finite.2012-12-24
  • 0
    May I know why should $\Phi(G)?2012-12-24
  • 0
    Since $G$ isn't abelian we can find two distinct maximal subgroups (each of one is abelian), say $U1, U2$. Then $=G$. The intersection of $U1$ and $U2$ is central. Am I wrong?2012-12-24
  • 0
    $G$ is a finite $p$-group, so it is obviously nilpotent (of class 2, in this case). The center and the commutator subgroup of $Q_8$ are the same, $\{1,-1\}$, so it's true for it.2012-12-24
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    Sorry, I missed the "equals" above :D2012-12-24
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    I only want to make sure: did you mean to write, one line before the last one,$$"\text{Then}\,\,\left(G'\leq\Phi(G)\leq Z(G)\right)\,\,H\triangleleft G"\,\,?$$ And if you did, why $\,H\triangleleft G\,$?2012-12-24
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    Yes. Something wrong? $H, then $H\vartriangleleft G$ (subgroups of the center are trivially normal)2012-12-24
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    It just looked funny: two lines before the end the line *ends* with a left parentheses "(", so I all the time thought you meant $\,\Phi(G)\leq Z(G)H\,$...the right, closing, parentheses looked like a typo. Perhaps it'd be a good idea to jump a line there.2012-12-24

2 Answers 2

3

I was thinking... $G$ is a finite, nilpotent (so also soluble) group; so there exist a $G_1\vartriangleleft G$ such that $|G:G_1|=p$. $G$ is minimal non abelian, hence $G=$. We can suppose $y\notin G_1$, but then there exist a $g\in G_1$ such that $y^n=xg$; so $G=$. Then, as above, $G'=[G_1,y]$ and every $x\in G'$ is a product of element of the form $[y^{n_1}g^{m_1}...y^{n_t}g^{m_t}, y^c]$. Hence every $x\in G'$ has the form: $[g^m, y^s]=[g, y]^{k}$.
$G'$ is cyclic and elementary abelian, since $G$ is Frattini-in-center, so $|G'|=p$.

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Have a look at page $6$ of Miller and Moreno.

  • 0
    "As this subgroup and $s$ must generate $G$, it follows that the commutator subgroup of $G$ is of order $p$". "this subgroup" referes to $G_1$? In such a case: $[G, G]\leq [G_1 , s]$ (since $G$ is nilpotent). But then? How can we use the p-group-fact?2012-12-25