On Frattini Subgroups (6.1.21-25)

Dummit and Foote Abstract Algebra, section 6.1, exercises 21-25:

21. Prove $\Phi(G)~\text{char}~G$.
22. Prove $N \trianglelefteq G ⇒ \Phi(N)≤\Phi(G)$. Give an explicit example where this containment does not hold when $N \not \trianglelefteq G$.
23. Compute $\Phi(S_3)$,$\Phi(A_4)$,$\Phi(S_4)$,$\Phi(A_5)$, and $\Phi(S_5)$.
24. Define $x~\text{is a nongenerator}~⇔∀H < G~[\langle~x,~H~\rangle ≠ G]$. Prove $\Phi(G)= \{x~|~x~\text{is a nongenerator}~\}$ when $G ≠ 1$.
25. Let $|G|< \infty$. Prove $\Phi(G)$ is nilpotent. [Use Frattini's Argument to show $∀P∈Syl(\Phi(G))~[P \trianglelefteq \Phi(G)]$]

Proof: Let $\mathcal{M}=\{...,H_i,...\}$ be the (potentially infinite, even uncountable) set of maximal subgroups of $G$, and $I$ be its indexing set.

Lemma 1: For any $φ∈Aut(G)$, we have $φ$ permutes the elements of $\mathcal{M}$. Proof: Pick an arbitrary $H_i∈\mathcal{M}$. If $φ(H_i)$ is not a maximal subgroup of $G$, then observe $φ(H_i) < K < G$. We have $H_i = φ^{-1}(φ(H_i)) < φ^{-1}(K)$, the last term of which is properly contained in $G$ (else $K=φ(G)=G$), therefore $H_i < φ^{-1}(K) < G$, a contradiction. Therefore $φ(H_i)$ is a maximal subgroup, and this action of $φ$ on $\mathcal{M}$ is well defined, injective by nature, and surjective since $φ^{-1}(H_i)$ will map to $H_i$. $\square$

Lemma 2: $φ(\bigcap_{i∈I}H_{i})=\bigcap_{i∈I}φ(H_i)$ for any isomorphism $φ$ of $G$, with $H_i≤G$ for all $i∈I$. Proof: $$φ(x)∈φ(\bigcap_{i∈I}H_{i})⇔x∈\bigcap_{i∈I}H_{i}⇔∀i∈I[x∈H_i]⇔$$ $$∀i∈I[φ(x)∈φ(H_i)]⇔φ(x)∈\bigcap_{i∈I}φ(H_{i})~~~\square$$

(21) Let $\pi$ denote the permutation of the indices of the elements of $\mathcal{M}$ by $φ∈Aut(G)$ by lemma 1. We have $$φ(\Phi(G))=φ(\bigcap_{i∈I}H_{i})=\bigcap_{i∈I}φ(H_{i})=\bigcap_{i∈I}H_{\pi(i)}=\bigcap_{i∈I}H_{i}=\Phi(G)$$
since $\pi(I)=I$.$~\square$

(22) (With aid from Project Crazy Project) (Assuming $G$ is finite) Lemma 3: Let $A ≤ C ≤ G$ and $B ≤ G$. Then $A(B∩C)=AB∩C$. Proof: ($\subseteq$) $x∈A(B∩C)⇒x=ad$ for some $a∈A$ and $d∈B∩C$, so that $ad∈AB$ and $ad∈C$, ergo $x∈AB∩C$. ($\supseteq$) $x∈AB∩C⇒x=ab \land x∈C$, so that $b=a^{-1}x∈C$ and now $b∈B∩C$, hence $x∈A(B∩C)$. $~\square$

Lemma 4: $\Phi(G)H < G$ for any $H < G$. Proof: Quickly done, this is a consequence of exercise 24. Proving it otherwise is a simple feat nonetheless.$~\square$

Now, we have $\Phi(N)~\text{char}~N \trianglelefteq G⇒\Phi(N) \trianglelefteq G$, so that $\Phi(N)K$ is a subgroup for any $K≤G$. Observe the implications: $$\Phi(N) \not ≤ \Phi(G )⇒ ∃i∈I[\Phi(N) \not ≤ H_i]⇒N \not ≤ H_i ⇒ H_i∩N < N ⇒$$ $$\Phi(N)(H_i∩N) < N ⇒ \Phi(N)H_i∩N < N ⇒ G∩N < N ⇒ N < N$$ We proceed to constructing a counterexample when $N \not \trianglelefteq G$. This lemma will be useful for the next exercise as well: Lemma 5: $\Phi(S_n)=\Phi(A_n)=1$ when $n \geq 5$. Proof: $\Phi(A_n) \triangleleft A_n$, so we must have $\Phi(A_n)=1$. As computed in 4.6.2, the only proper normal subgroups of $S_n$ are $A_n$ and $1$. Since $A_n$ is a maximal subgroup and not the only one (place $\langle~(1 2)~\rangle$ in a maximal subgroup, for instance), we must have $\Phi(S_n)=1$.$~\square$

Now, associate $Q_8$ with its isomorphic image in $S_8$. By order considerations, it is not equal to $S_8$,$A_8$, or $1$, so it is nonnormal. We can see $\Phi(Q_8)=\langle~-1~\rangle$ is nontrivial, and by the above lemma $\Phi(S_8)$ is trivial, so the containment doesn't hold.$~\square$

(23) The groups of order 3 and 2 by Cauchy are maximal by their orders, and have trivial intersection, so $\Phi(S_3)=1$. By 3.5.8, $\langle~(1 2)(3 4),~(1 3)(2 4)~\rangle$ and $\langle~(1 2 3)~\rangle$ are maximal subgroups, and have trivial intersection, therefore $\Phi(A_4)=1$.

For $\Phi(S_4)$, take an arbitrary subgroup $M$ of order 12 in $S_4$. By Cauchy, $M$ has an element of order 3, which must be a 3-cycle. Since conjugation of $M$ by elements of $S_4$ are automorphisms of $M$ (since $M \trianglelefteq S_4$ by order), we have that $M$ contains all 3-cycles, and is thus equal to $A_4$. Now, $\langle~(1 2),(1 2 3)~\rangle$ can only contain elements of $S_3$ due to its generators fixing 4, and indeed it does generate at least and thus exactly 6 elements. Therefore it is maximal since otherwise it is contained in a group of order 12, which must be $A_4$ by above, a contradiction since $A_4$ doesn't contain a subgroup of order 6 by 3.5.8. By this, $\Phi(S_4)$ is either of order 1 or 3, and not the latter since $\Phi(S_4) \trianglelefteq S_4$ would imply $\Phi(S_4)$ contains every 3-cycle.

By the lemma, $\Phi(A_5)=\Phi(S_5)=1$.$~\square$

(24) (Assuming $G$ is finite)($\subseteq$) Let $x∈\Phi(G)$, and for any $H < G$, place $H$ in a maximal subgroup $H^*$. We have $x∈H^*$, therefore $\langle~x,~H~\rangle ≤ \langle~x,~H^*~\rangle = H^* < G$. ($\supseteq$) Let $x$ be a nongenerator, so that for any maximal subgroup $H^*$, we must have $\langle~x,~H^*~\rangle = H^*$, so that $x∈H^*$.$~\square$

(25) Let $P$ be an arbitrary Sylow subgroup of $\Phi(G)$. By Frattini, we have $\Phi(G)N_G(P)=G$. Since $\Phi(G)N_G(P)$ is clearly not a proper subgroup, we must have $N_G(P)$ is not a proper subgroup by lemma 4, which is to say $N_G(P)=G$ and now $P \trianglelefteq G$, and in particular $P \trianglelefteq \Phi(G)$, so that $\Phi(G)$ is nilpotent by theorem 3(3).$~\square$