Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

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

share|cite|improve this question
We see this nice deduction when $|G|=p^3$ and the group is non-abelian and finite. – Babak S. Dec 24 '12 at 16:46
May I know why should $\Phi(G)<Z(G)$? – Babak S. 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.

share|cite|improve this answer
"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
up vote 3 down vote accepted

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 $|G'|=p$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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