Suppose $p$ is a prime number and $G$ is a finite group, such that $\Phi(G) = D_4 = \langle a \rangle_4 \rtimes \langle b \rangle_2$, where $\Phi$ denotes the Frattini subgroup. Is it always true, that $32$ divides $|G|$?

To solve that problem I tried to apply the method from the answer to the following question: A question about Frattini subgroup of specific form

That is, what I got:

Let $\Phi(G) = D_4$, and $\Phi(G) \le P \in {\rm Syl}_2(G)$.

Now $\Phi(G)$ cannot have a complement in $G$, since otherwise that complement would be contained in a maximal subgroup that did not contain $\Phi(G)$. So by the theorem of Gaschütz, $\Phi(G)$ does not have a complement in $P$. So $\Phi(G) < P$, and we only have to consider the case when $P=16$. Then, elements $g \in P \setminus N$ must have order dividing $8$, with $g^2 \in \Phi(G)$.

Now the conjugation action of $G$ on $\Phi(G)$ induces a subgroup $\bar{G} = \frac{G}{C_G(\Phi(G))}$ of ${\rm Aut}(\Phi(G)) \cong D_4$. If the image $\bar{P}$ of $P$ in $\bar{G}$ is not normal in $\bar{G}$, then $\bar{G}$ has more than one Sylow $2$-subgroup. But any subgroup of $D_4$ is a $2$-group, and thus $\bar{G}$ has a unique Sylow $2$-subgroup.

So $\bar{P} \unlhd \bar{G}$. Suppose $M = \langle g^2 \mid g \in P \rangle$. $M$ is a characteristic subgroup of $G$ and is contained in $\Phi(G)$. If $|P| \leq 16$, then there are three cases:

First one is, when $M$ is the proper characteristic subgroup of $\Phi(G)$ of order $4$. Then the image $\frac{\Phi(G)}{M}$ of $\Phi(G)$ has a complement in $\frac{P}{M}$, and hence, by Gaschütz's theorem again, $\frac{\Phi(G)}{M}$ has a complement $\frac{H}{M}$ in $\frac{G}{M}$. Then $|G:H| = 2$. So $H$ is a maximal subgroup of $G$ not containing $\Phi(G)$, contradiction.

Second one is, when $M$ is the proper characteristic subgroup of $\Phi(G)$ of order $2$.

Third one is, when $M=\Phi(G)$.

And here I am stuck, not knowing, what to do with the second and the third cases.


You are just trying to rule out the case when $|P|=16$. In that case $G/\Phi(G)$ has twice odd order, so if has a normal subgroup $M/\Phi(G)$ of index $2$.

By Schur-Zassenhaus, $\Phi(G)$ has a complement $N$ in $M$. Since ${\rm Aut}(\Phi(G)) = {\rm Aut}(D_4)$ is a $2$-group, $N$ must centralize $\Phi(G)$, so $M = \Phi(G) \times N$, and hence $N$ is characteristic in $M$ and so normal in $G$.

But now if we let $Q$ be a maximal subgroup of $P$ other than $\Phi(G)$, then $QM$ is a maximal subgroup of $G$ that does not contain $\Phi(G)$, contradiction.


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