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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 $k$, $1<k<n-1$?

  • 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.

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It would be better to ask single questions. –  Phira Oct 8 '11 at 11:12
    
I deleted my answer because there is something strange. Can you please define "non-generator"? –  Phira Oct 8 '11 at 11:17
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@Phira: Thank you for reminding me. I thought the questions have something in common, so I put them together. Maybe I shouldn't have done like this. A nongenerator for a group $G$ is an element $a$, such that for any subset $X$ of $G$, if $G = <a,X>$, then $G = <X>$. I thought this was a commonly used definition, so I omitted it. I will add this definition to my question. –  ShinyaSakai Oct 8 '11 at 11:32
    
Regarding 1.: If $n$ is even, then $\langle a^2,b\rangle$ is of index 2 and hence maximal. The intersection $\langle a^2,b\rangle$ and $\langle a\rangle$ is $\langle a^2\rangle$ and the quotient is $C_2\times C_2$. You can easily show that in fact, $\Phi(D_{n})=\langle a^2\rangle$. –  Alex B. Oct 8 '11 at 12:06
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If $d$ is a divisor of $n$ other than 1 or $n$, then $\langle b, a^d \rangle$ is a proper subgroup of $D_n$ (isomorphic to $D_{n/d}$). So $\{1,b\}$ isn't a maximal subgroup of $D_n$, unless $n$ is prime. –  Ted Oct 8 '11 at 17:49

2 Answers 2

up vote 4 down vote accepted

Regarding $S_n$: try to prove that the subgroup of permutations which fix a particular point is a maximal subgroup of $S_n$. These subgroups obviously intersect trivially, so the Frattini subgroup is trivial.

Regarding the third question: since the index of $M$ in $G$ is $p$, the factor group is isomorphic to $C_p$ and, in particular, is abelian. But you know that $G'$ is the smallest normal subgroup in $G$ for which the quotient group is abelian.

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Thank you very much for the answer. I am sorry but I am still confused on the third question. I don't think every group $G$ with each maximal subgroup of prime index satisfies $G' \subseteq \mathrm{Frat}G$: for the group $S_3$, every maximal subgroup is of primed index, but as it is not commutative, $S_3' \subsetneq \mathrm{Frat}S_3 = 1$... I am confused with the taking factor part, because $M$ is not normal. Thanks again. –  ShinyaSakai Oct 9 '11 at 14:12
    
As you state in your question, maximal subgroups of nilpotent groups are normal, so it makes sense to talk about the quotient. Regarding your $S_3$ example, let me clarify: a normal subgroup $N$ of a group $G$ contains $G'$ iff $G/N$ is abelian. $S_3$ has non-normal maximal subgroups (they are of prime index but that doesn't matter); taking their normal closure would make them contain $S_3'$, but it would also enlarge them to the whole $S_3$. –  Miha Habič Oct 9 '11 at 15:41
    
I see... I neglected the normalizer condition... Thank you very much. –  ShinyaSakai Oct 9 '11 at 16:20

If $1 \le d < n$ and $d$ is a divisor of $n$, then $\langle b, a^d \rangle$ is a proper subgroup of $D_n$. Which of these subgroups are maximal? What do you get when you intersect the maximal ones?

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Thank you very much for the enlightening answer~ If $n = p_1^{k_1} \cdots p_l^{k_l}$ for primes $p_i$ and positive integers $k_i$. Then the maximal subgroups of $D_n$ are $\langle a \rangle$, and $\langle a^{p_i}, b \rangle$. So, $\mathrm{Frat}D_n = \langle a^{p_1 \cdots p_l} \rangle$. Is this right? –  ShinyaSakai Oct 9 '11 at 13:07
    
I think I was wrong. There are other types of maximal subgroups. For example, when $n$ is a prime, $\{ 1, ab \}$ is a maximal subgroup... –  ShinyaSakai Oct 9 '11 at 13:26
    
There are indeed other maximal subgroups, but you're very close. Assume $n$ is odd for the moment; then all maximal subgroups are conjugate to the ones you have already found. This is because all reflections are conjugate in $D_n$, so if a maximal subgroup of $D_n$ does not consist entirely of rotations, then some conjugate of it contains $b$, and this leads to the subgroup $\langle b, a^{p_i} \rangle$. Since $\langle a^{p_1 \ldots p_l} \rangle$ is already normal, intersecting by these conjugates will not give us a smaller group. For $n$ even, it's only a little more complicated. –  Ted Oct 10 '11 at 2:04

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