# Nonabelian $p$-groups all of whose proper subgroups are abelian.

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 $1<H<G'$.
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
$G'$ is elementary abelian since $G$ is Frattini-in-center.

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We see this nice deduction when $|G|=p^3$ and the group is non-abelian and finite. – S. Snape Dec 24 '12 at 16:46
May I know why should $\Phi(G)<Z(G)$? – S. Snape Dec 24 '12 at 16:56
Since $G$ isn't abelian we can find two distinct maximal subgroups (each of one is abelian), say $U1, U2$. Then $<U1,U2>=G$. The intersection of $U1$ and $U2$ is central. Am I wrong? – W4cc0 Dec 24 '12 at 16:58
Sorry, I missed the "equals" above :D – W4cc0 Dec 24 '12 at 17:10
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. – DonAntonio Dec 24 '12 at 17:52

Have a look at page $6$ of Miller and Moreno.
"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? – W4cc0 Dec 25 '12 at 11:29
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 $G=<x, y>$. We can suppose $y\notin G_1$, but then there exist a $g\in G_1$ such that $y^n=xg$; so $G=<y,g>$. 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 $|G'|=p$.