8
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ We see this nice deduction when $|G|=p^3$ and the group is non-abelian and finite. $\endgroup$ – mrs Dec 24 '12 at 16:46
  • $\begingroup$ May I know why should $\Phi(G)<Z(G)$? $\endgroup$ – mrs Dec 24 '12 at 16:56
  • $\begingroup$ 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? $\endgroup$ – W4cc0 Dec 24 '12 at 16:58
  • 1
    $\begingroup$ Sorry, I missed the "equals" above :D $\endgroup$ – W4cc0 Dec 24 '12 at 17:10
  • 1
    $\begingroup$ 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. $\endgroup$ – DonAntonio Dec 24 '12 at 17:52
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ "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? $\endgroup$ – W4cc0 Dec 25 '12 at 11:29
3
$\begingroup$

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 $|G'|=p$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.