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Let $G$ be a finite $p$-group with has more than one maximal subgroup. Prove that $G$ has at least $p+1$ maximal subgroups.

I don't have idea. Help me.

Thanks in advanced.

EDIT: I found a result that group of order $p^2$ has exactly $p+1$ maximal subgroup (see page $27$ of Group Theory I - Michio Suzuki). So for a finite $p$-group we are done that it has at least $p+1$ maximal subgroups. I hope I don't have mistakes. :)

user111636
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    Dear user111636 this appears to be the fifth post that consists of a problem statement with no effort in the last 12 hours. This is likely to attract negative attention! Really, all it takes is minimal effort to include where you're stuck and you will get much better help. – rschwieb Jan 04 '14 at 13:29

2 Answers2

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Hint: The intersection of all maximal subgroups $F$ is called a Frattini subgroup. It is known that $G/F$ is Abelian (since $G$ is nilpotent). So it is sufficient to consider maximal subgroups of the Abelian group $G/F$.

Boris Novikov
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    In fact $G/F$ is elementary abelian, so the number of maximal subgroups is exactly $(p^k-1)/(p-1)$, where $|G/F|=p^k$. – Derek Holt Jan 04 '14 at 10:40
  • @Derek Holt: Of course, I gave only a hint. :-) – Boris Novikov Jan 04 '14 at 10:46
  • it is the Frattini subgroup of $;G;$ , denoted by $;\Phi(G);$ , and in fact $;\Phi(G)=G'G^p;$ , which shows almost immediately what Derek mentions. Thus, $;G/\Phi(G);$ is an elementary abelian $;p-$group and thus a vector space over the prime field of characteristic $;p;$. The number in Derek's comment follows at once even from basic linear algebra. – DonAntonio Jan 04 '14 at 11:29
  • @DonAntonio: Sorry for "$a$ Frattini" - my English is bad. :-( – Boris Novikov Jan 04 '14 at 12:35
  • I know "the" and "a" are difficult for people whose native languages do not have articles, but the difference is imporatant in mathematics, because "the" implies uniqueness, whereas "a" suggests non-uniqueness. – Derek Holt Jan 04 '14 at 14:20
  • @Derek Holt: Oh, I know this, but I read in some textbook that after "is called" one writes usually "a". – Boris Novikov Jan 04 '14 at 14:33
  • @user111636: You are welcome. – Boris Novikov Jan 05 '14 at 07:13
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An elementary solution, which does not require any knowledge of the Frattini subgroup, is the following.

Let $M$ and $N$ be two distinct maximal subgroups. They are known to be normal in $G$, and the homomorphism $$ G \to G/M \times G/N, \qquad x \mapsto (xM, xN), $$ has kernel $K = M \cap N$, of index $p^2$ in $G$, so $G/K \cong G/M \times G/N$ is elementary abelian of order $p^2$, and its $p+1$ subgroups of index $p$ correspond to $p+1$ distinct maximal subgroups of $G$.

Andreas Caranti
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  • Can you explain why $G/K$ has $p+1$ subgroups? – user111636 Jan 05 '14 at 00:48
  • @user111636, an elementary abelian $p$-group (that is, an abelian group in which all elements $a$ satisfy $a^p = 1$, where $p$ is a prime) is the same thing as a vector space over the field $F$ with $p$ elements. So $G/K$ is a two-dimensional vector space over $F$. If $e, f$ is a basis, the subspaces of dimension $1$ are those generated by $a$ and the $x a + b$, for $x \in F$. – Andreas Caranti Jan 05 '14 at 11:13