Frattini Subgroup

Definition

Let $G$ be a finite group. The Frattini subgroup of $G$, denoted by $\Phi (G)$, is the intersection of all maximal subgroups of $G$.

For the case that $G$ has no maximal subgroups, it is defined by $\Phi (G)=G$.

It is easy to prove that $\Phi (G)$ is a characteristic subgroup. Moreover, some properties are given as follows.

Set of Nongenerators

Definition

An element $x\in G$ is called a non-generator if $⟨x,S⟩=G$, then $⟨S⟩=G$ for any $S\subseteq G$.

Examples.

• Let $G=C_{p^2}$. Then for any $a\in G$, $a^p$ is a non-generator.
• $Q_8$ has non-generator s of order $1$ or $2$.
• The set of non-generators of $D_{2n}$ is $\{a^m,a^{m}b:(m,n)\ne 1\}$.
• Each non-identity element of a non-abelian simple group is not a non-generator element. In fact, each non-abelian simple group can be generated by “1.5” elements. That is, for a non-abelian simple group $G$ and any $1\ne g\in G$, there exists a $h\in G$ such that $G=⟨g,h⟩$.

Theorem

$\Phi (G)$ is the set of non-generators of $G$.

\begin{proof} Define $X$ as the set of all non-generators of $G$. For any $x\in X$ and any maximal subgroup $M$, $⟨M,x⟩=⟨M⟩$ yields $x\in M$. Thus $X\subseteq M$ for any $M$ and so $X\subseteq \Phi (G)$.

Conversely, assume that there exists a $g\in \Phi (G)\setminus X$. Then there is a subset $S$ such that $⟨S⟩<G$ and $⟨S,g⟩=G$. Let $H=⟨S⟩$. If $H$ is not a maximal subgroup, there is a maximal subgroup $K$ with $H<K<G$. Since $g\in \Phi (G)⩽K$, we have $K⩾⟨H,g⟩=G$, which is impossible. Hence, $H$ is a maximal subgroup and so $g\in H$. Then $⟨H,g⟩=H\ne G$, which is a contradiction. Therefore, we have $X\supseteq \Phi (G)$. \end{proof}

Theorem

For $N⊲G$ and $H⩽G$, if $N⊲\Phi (H)$, then $N⩽\Phi (G)$.

\begin{proof} Assume that $N/⩽\Phi (G)$, then there exists a maximal subgroup $M<G$ such that $N/⩽M$. It deduces that $G=NM$ and so $H=H\cap NM=N(M\cap H)$. Since $N⊲\Phi (H)$, there is $H=M\cap H$ and so $M⩾H⩾\Phi (H)⩾N$, which is impossible. \end{proof}

Corollary

If $N⊲G$, then $\Phi (N)⩽\Phi (G)$.

\begin{proof} It is a direct corollary of ^b50be5, and the following proof comes from Group Theory - 2024spring.

Since $\Phi (N){⊲}_{\mathrm{char}}N$, we have that $\Phi (N)⊲G$. It suffices to show that for any maximal subgroup $M<G$, there is $M⩾\Phi (N)$. Otherwise, if there exists a maximal subgroup $M$ such that $\Phi (N)⊈M$, then $⟨M,\Phi (N)⟩=M\Phi (N)=G$. Note that

$$G/N\cong M\Phi (N)N/N\cong M\Phi (N)/(M\cap N)\Phi (N)\cong G/(M\cap N)\Phi (N).$$

As $N⩾(M\cap N)\Phi (N)$, it yields that $N=(M\cap N)\Phi (N)$ and so $N=⟨M\cap N,\Phi (N)⟩=M\cap N$, contradiction. \end{proof}

Remark. In general, $K⩽G$ does not always yield $\Phi (K)⩽\Phi (G)$. For example, take $K=D_8$ and $G=S_5$. (Remark that $\Phi (S_n)=\{e\}$, because $S_{n-1}⩽S_n$ is a maximal subgroup.)

Nilpotent Subgroups

Gaschutz

Let $D,N⊲G$ such that $D⩽\Phi (G)\cap N$. Then $N/D$ is nilpotent iff $N$ is nilpotent.

\begin{proof} Assume that $N/D$ is nilpotent. Let $P$ be a Sylow $p$-subgroup of $N$. Then $PD/D$ is a Sylow $p$-subgroup of $N/D$, and

$$PD/D{⊲}_{\mathrm{char}}N/D⊲G/D.$$

Hence, $PD/D⊲G/D$ and $PD⊲G$. It follows that $PD⊲N$. By Frattini’s argument, we have $N=PD{N}_N(P)=P{N}_N(P)=N_N(P)$, so $P⊲N$. Therefore, $N$ is nilpotent. \end{proof}

Corollary

• (Frattini theorem) $\Phi (G)$ is nilpotent.
• $G$ is nilpotent iff $G/\Phi (G)$ is nilpotent.
• (Wielandt) $G$ is nilpotent iff $G^{\prime }⩽\Phi (G)$.

\begin{proof} (i) Taking $N=D=\Phi (G)$. Then by ^124c42 $N$ is nilpotent.

(ii) Let $N=G$ and $D=\Phi (G)$. Then $N/D=G/\Phi (G)$ is nilpotent iff $N=G$ is nilpotent, by ^124c42.

(iii) Suppose $G$ is nilpotent. Let $M$ be a maximal subgroup of $G$. Then $M⊲G$ and $G/M\cong {\mathbb{Z}}_p$ is abelian by ^9gfd7n. Thus $M⩾G^{\prime }$. By the arbitrary of $M$, we have that $G^{\prime }⩽\Phi (G)$. Conversely, assume that $G^{\prime }⩽\Phi (G)$. Then $G/\Phi (G)$ is abelian and so is nilpotent. Then by (ii), $G$ is nilpotent. \end{proof}

Exactly Elementary Abelian

Theorem

If G is a finite $p$-group, then $\Phi (G)$ is the smallest normal subgroup such that its quotient is elementary abelian. Moreover, if the quotient group $G/\Phi (G)$ has order $p^k$, then $k$ is the smallest number of generators for $G$.

\begin{proof} Since $G$ is a $p$-group, then $G$ is nilpotent. Thus $\Phi (G)⩾G^{\prime }$ by ^395831 (iii), and so $G/\Phi (G)$ is abelian. If there is a $\overline{g}\in G/\Phi (G)$ with $|\overline{g}|$ is not a prime, then ${\overline{g}}^a$ is a non-generator with $|\overline{g}|=ab$, which contradicts with ^7vww3g. Therefore, $G/\Phi (G)$ is elementary abelian.

Assume that $N⊲G$ is a normal subgroup of $G$ and $G/N$ is elementary abelian. It remains to show $N⩾\Phi (G)$. Suppose that $G/N=⟨\overline{g_1}⟩\times \cdots \times ⟨\overline{g_m}⟩\cong {\mathbb{Z}}_p^m$. For any preimage $g_i\in G$ of $\overline{g_i}$, $g_i$ is not a non-generator of $G$ because $⟨g_1,\cdots ,g_m,N⟩=G$ and $⟨g_1,\cdots ,{\check{g}}_i,\cdots ,g_m,N⟩\ne G$. Therefore, if $g\in G\setminus N$, then $g$ is not a non-generator and so $N⩾\Phi (G)$.

For the “moreover” part, since $G/\Phi (G)\cong {\mathbb{Z}}_p^k$ for some $k$, it can be seen as a linear space and it has a basis $\{{\overline{e}}_1,\cdots ,{\overline{e}}_k\}$. Let $\{e_1,\cdots ,e_k\}$ be a set of preimage of $G\to G/\Phi (G)$. Then $⟨e_1,\cdots ,e_k,\Phi (G)⟩=G$ and so $⟨e_1,\cdots ,e_k⟩=G$ by ^7vww3g. \end{proof}

Remark. It is similar as Nakayama lemma.

Preserved by Direct Product

Proposition

If H and K are finite groups, then $\Phi (H\times K)=\Phi (H)\times \Phi (K)$.

\begin{proof} If $H_0$ and $K_0$ are maximal subgroups of $H$ and $K$, respectively, then $H_0\times K$ and $H\times K_0$ are maximal subgroups of $H\times K$ and so $H_0\times K_0\subseteq (H_0\times K)\cap (H\times K_0)$. Therefore, we have $\Phi (H)\times \Phi (K)\subseteq \Phi (H\times K)$.

Conversely, suppose that $L$ is a maximal subgroup of $H\times K$. Define ${\pi }_1:H\times K\to H$ and ${\pi }_2:H\times K\to K$ be the canonical quotient maps. For any $h\in H\setminus {\pi }_1(H)$, we have $(h,1)\notin H\times K$ and so $⟨L,(h,1)⟩=H\times K$. Therefore, $L_H:={\pi }_1(L)$ is a maximal subgroup of $H$. Similarly, $L_K:={\pi }_2(L)$ is a maximal subgroup of $K$. Then we have $\Phi (H)\times \Phi (K)\supseteq \Phi (H\times K)$. Now we finish the proof. \end{proof}