Frattini Subgroup of p-Groups (6.1.26a-b)

Dummit and Foote Abstract Algebra, section 6.1, exercise 26a-b:

(a) Let the overbar denote passage into $P/\Phi(P)$. Prove $\overline{P}$ is an elementary abelian $p$-group. [Prove $P'≤\Phi(P)$ and $x^p∈\Phi(P)$ for any $x∈P$]

(b) Prove that if $P/N$ is elementary abelian, then $\Phi(P)≤N$.

Proof: (a) Note that $p$-groups are nilpotent, so that every maximal subgroup of $P$ is normal and of prime index by 6.1.4, so that the quotient group of $P$ over any maximal subgroup is abelian, so that by an extension of 5.4.14, $P'$ is contained within the intersection of all maximal subgroups, which is to say $P'≤\Phi(P)$. Similarly, since the quotient groups are of prime order, we have $\overline{x}^p$ is trivial for any $x∈P$, which is to say $x^p∈\Phi(P)$.$~\square$

(b) Lemma 1: $\Phi(Z_p^t)=1$ for any prime $p$ and nonnegative integer $t$. Proof: Let $(x_1,...,x_r,...,x_t)$ be any nontrivial element with a nontrivial coordinate designated by $x_r$. We have $\langle~e_1,...,~e_{r-1},~e_{r+1},...,~e_t~\rangle$ is a subgroup of order $p^{t-1}$ that is maximal by its index, and this particular subgroup doesn't contain $(x_1,...,x_r,...,x_t)$. Since the nontrivial element was arbitrary, we have $\Phi(Z_p^t)=1$.$~\square$

Since $P/N$ is elementary abelian, we have the intersection of its maximal subgroups is the identity, which is to say (by the Lattice Isomorphism Theorem) that the intersection of the set of maximal subgroups of $P$ containing $N$ is contained in $N$. By definition, $N$ is contained within the intersection of the set of maximal subgroups of $P$ containing $N$, so there is an equality. Since $\Phi(P)$ is contained in all maximal subgroups of $P$, we have $\Phi(P)≤N$.

In other words, if $φ : P → E$ is any homomorphism of $P$ into an elementary abelian group $E$, then $φ$ factors through $\Phi(P)$. $$\begin{array}{ccc} P&\stackrel{\pi}{\longrightarrow}&{P/\Phi(P)}\\ &\searrow\scriptstyle{φ}&\downarrow\scriptstyle{ψ}\\ &&E \end{array}$$ $~\square$