According to Derek J.S Robinson's A Course in the Theory of Groups, the Frattini subgroup of a group $G$, denoted $\mathrm{Frat}G$ is defined to be the intersection of all maximal subgroups of $G$. When $G$ has no maximal subgroup, $\mathrm{Frat}G$ is set to be $G$ itself. It can be proved that $\mathrm{Frat}G$ is in fact the set of all nongenerators of $G$. (Here, a nongenerator for a group $G$ is an element $a$, such that for any subset $X$ of $G$, if $G = \langle a,X\rangle$, then $G = \langle X\rangle$.) The definition and property can be applied to find the Frattini subgroups of a given group, but I still have problems in doing this.
- Find the Frattini subgroup of $D_n$, the dihedral group of order $2n$.
$D_n = \langle a,b | a^n = b^2 =1, abab =1\rangle$. The subgroup $\langle a\rangle = \{ 1, a, a^2, \cdots, a^{n-1} \}$ is maximal in $D_n$, so $\mathrm{Frat}D_n \subseteq \langle a\rangle$. If $n$ is odd, then $\langle b\rangle = \{ 1, b \}$ is also a maximal subgroup of $D_n$. $\langle a\rangle \cap \langle b\rangle = 1$, so $\mathrm{Frat}D_n =1$. But what will happen if $n$ is even? Do I have to test whether $a^k$ is a nongenerator for any integer $k$, $1
- Prove that the Frattini subgroup of $S_n$ is trivial.
I think of the proof in these two directions: no nontrivial element of $S_n$ is contained in every maximal subgroup of $S_n$, and, no nontrivial element of $S_n$ is a nongenerator. As $A_n$ is a maximal subgroup, $\mathrm{Frat}S_n \subseteq A_n$. Moreover, the Frattini subgroup is normal in $S_n$, so if $n \geq 5$, this subgroup must be $A_n$ or trivial. If for any given nontrivial element of $A_n$, I could find some maximal subgroup of $S_n$ not containing it, then I could prove the result. But I don't know how to find them.
Let $G$ be a finite $p$-group. Then \mathrm{Frat}G = G'G^p.
In the proof, the author said:
As $G$ is nilpotent and satisfies the normalizer condition, we have $M \lhd G$ for a maximal subgroup $M$ of $G$. Moreover, $|G:M| = p$. Hence G'G^p \leq \mathrm{Frat}G.
I don't know why this is true. It is easy to see that $G^p \leq \mathrm{Frat}G$, because $G^p$ is contained in every maximal subgroup of $G$. But why is G' contained in $ \mathrm{Frat}G$? This is not true in general, for example, let $G = S_3$, then a \in G' but $a \notin \mathrm{Frat}G$. Is it true for all the $p$-groups? all the finite $p$-groups? Why is it true?
Thank you very much.