Frattini sugroup and normal subgroup

Question

For any group $G$, let $\Phi(G)$ denote the Frattini subgroup of $G$.

Let $G$ be a finite group, such that $\dfrac{G}{\Phi(G) } \cong K \times \mathbb{Z}_{p}$, where $p$ is prime number.

Prove that if $ p \nmid |K|$ then there exists a normal subgroup $L$ of $G$, ($ L \trianglelefteq G$) such that for some $n \in \mathbb{N}$, $G \cong L \times \mathbb{Z}_{p^n}$, and $\dfrac{L}{\Phi(L) } \cong K$.

I think we have: for a finite group $G$, $|G/\Phi(G)|$ is divisible by all primes $p$ dividing $|G|$. It uses the Schur-Zassenhaus Theorem.

First note that each Sylow $p$-subgroup $P$ of $\Phi(G)$ is normal in $G$. (So, in particular, $\Phi(G)$ is nilpotent.) To see that, we have $G = \Phi(G)N_G(P)$ by the Frattini Argument, and then by the fact that $\Phi(G)$ consists of the non-generators of $G$, we have $G = N_G(P)$.

Now if there is a prime $p$ dividing $|G|$ but not dividing $|G/\Phi(G)|$, then $\Phi(G)$ contains a Sylow $p$-subgroup $P$ of $G$ and $P \unlhd G$. So, by the Schur-Zassenhaus Theorem, $P$ has a complement $H$ in $G$. Let $M$ be a maximal subgroup of $G$ containing $H$. Then $p$ divides $|G:M|$ and hence $p$ divides $|G/\Phi(G)|$, contradiction.

Answer 1

I will write $G/\Phi(G) \cong K \times P$, with $P \cong C_p$; i.e. $P$ is cyclic of prime order $p$.

Now $\Phi(G)$ is nilpotent, so we have $\Phi(G) = M \times Q$ with $M$ a $p'$-group and $Q$ a $p$-group. (A finite $p'$-group is defined to be one in which the order is not divisible by the prime $p$.)

Let $J$ be the inverse image of $K$ in $G$. Then $J$ has the normal Sylow $p$-subgroup $Q$ and so, by the Schur-Zassenhaus theorem, it has a $p$-complement $L$ (i.e. $QL=J$ and $Q \cap L = \{1\}$), and all $p$-complements of $Q$ in $J$ are conjugate in $J$.

Since $M$ is a normal $p'$-subgroup of $J$, we have $M \le L$ and $L/M \cong K$. Note that $L$ is also a $p$-complement in $G$.

Now applying the Frattini argument to the $p$-complement $L$ in the normal subgroup $J = LQ$ of $G$, we get $G = N_G(L)LQ = N_G(L)Q$, and since $Q \le \Phi(G)$, this gives $G=N_G(L)$, so $L \lhd G$.

Similarly, let $S$ be the inverse image of $P$ in $G$, and let $R \in {\rm Syl_p}(S)$ (and hence $R \in {\rm Syl}_p(G)$). So $Q < R$, with $Q/R \cong P$. Applying the Frattini argument to the Sylow $p$-subgroup $R$ of the normal subgroup $S=RM$ of $G$, we have $G = N_G(R)RM=N_G(R)M$, and since $M \le \Phi(G)$, this implies that $N_G(R) = R$, and so $R \lhd G$.

So we now have $G=LR$ with $L \cap R = \{1\}$, and hence $G = L \times R$ which (by a standard result) implies that $\Phi(G) = \Phi(L) \times \Phi(R)$, and so $\Phi(L)=K$ and $\Phi(R)=Q$.

So $L/\Phi(L) \cong K$ and $R/\Phi(R) \cong P$, which proves the result. By a standard result this imples that $R$ is cyclic, which completes the proof.

Answer 2

Let $P \in {\rm Syl}_p(G)$ . we show that $P \lhd G$ and cyclic .we know $P\Phi(G) \lhd G$ now with Frattini argument we have : $G =\Phi(G) P N_G(P) = N_G(P)$ so $P \lhd G$ .

with Schur–Zassenhaus theorem , $P$ has a complement $L$ in $G$ then $G=LP$.let $\Phi(P) < \Phi(G) \cap P$ (proper subgroup) , then $P$ has a maximal subgroup $M$ such that $P=M (\Phi(G) \cap P)$ then $G=LM$ and we have a contradiction so $\Phi(P) = \Phi(G) \cap P$ and $P$ is cyclic because we have :$$\mathbb{Z_p}\cong \frac{P\phi(G)}{\phi(G)}\cong \frac{P}{\phi(G)\cap P}=\frac{P}{\phi(P)} $$ now we show $L \lhd G$ :

note that $[P,L] \le \phi(G)$ then $[P,L] \le \phi(P)$. by coprime action property $P=[P,L] C_p(L) $ so $P=C_p(L) $ and $[P,L]=1$ then $L \lhd G$.

finally $\Phi(L) = \Phi(G) \cap L$ and we have $ K \cong \frac{L}{\phi(L)}$