# A question about Frattini subgroup of specific form v2.0

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.