# Translation of a German paper

I'd like to have a translation (in English) of a paper of Klaus Doerk published in Journal of Algebra: http://www.sciencedirect.com/science/article/pii/S0021869384711999

It is 4,5 pages long with little mathematical notation and I am prepared to pay for this.

• No need to pay. We are all slaves here :) Nov 3, 2014 at 18:52
• Many thanks to both people who replied! A link to the full text: dropbox.com/s/bpzn237srlpz27m/… Nov 3, 2014 at 18:54
• Hopefully, Elsevier won't hit you too hard for this endeavour. Nov 3, 2014 at 19:43
• Oh, all right I have removed the dropbox file: libgen.org/scimag/… Nov 3, 2014 at 19:52
• Elsevier can take it up with libgen :) Nov 3, 2014 at 19:52

Here's what the preview allows me to see:

## On finite solvable groups that behave like nilpotent groups with respect to the Frattini subgroup.

Let $\Phi(G)$ denote the Frattini subgroup of a group $G$. If $G$ is a finite nilpotent group and $U\le G$ and $N\lhd G$, then it is known that $\Phi(U)\le \Phi(G)$ and $\Phi(G/N)=\Phi(G)N \big/ N$. The aim of this article is to characterize those finite solvable groups for which the Frattini group also has the described property. It proves very useful for this to use the language and theory of classes of groups as described e.g. in . All groups considered are assumed finite and solvable (unless mentioned otherwise).

Let $\mathfrak N$ denote the class of finite nilpotent groups, $\mathfrak S$ the class of finite solvable groups, and $\mathfrak E$ the class of finite groups. The closure operator $E_\Phi$ is defined as follows:

If $\mathfrak Z$ is a class of finite groups, let $$E_\Phi\mathfrak Z=\{\,G\mid G\text{ has a normal subgroup N with N\le \Phi(G) and G/N\in\mathfrak Z}\,\}.$$ (See also [1, II (1.6)]). A class $\mathfrak Z$ is called a formation if it is closed under taking epimorphic images and subdirect products. Local formations are defined in [1, IV, §3] and the elementary properties of Schunck classes can be found in [1, III, §2]. If $\mathfrak F$ is a formation, we denote with $G^{\mathfrak F}$ the smallest normal subgroup of $G$ with $G/G^{\mathfrak F}\in\mathfrak F$.

Definition. Let $\mathfrak X=\{\, G\in \mathfrak S\mid U\le V\le G\to \Phi(U)\le \Phi(V)\,\}$ and $$\mathfrak F=\{\,G\in\mathfrak S\mid \text{All Sylow subgroups of G are elementary abelian}\,\}.$$ The corresponding classes if one allows all finite groups for $G$ are denoted as $\overline {\mathfrak X}$ and $\overline { \mathfrak F}$. The classes $\mathfrak X$ and $\overline {\mathfrak X}$ are obviously closed under taking subgroups, and $\mathfrak F$ and $\overline {\mathfrak F}$ are formations closed under subgroups.

[End of first page; in looks in my copy (scan from the libgen link in comments) like some text might be missing here?]

Lemma 1. Let $G$ be a finite (not necessarily solvable) group, whose Sylow subgroups are all elementary abelian. Then $\Phi(G) = 1$. If $G$ is moreover solvable, then the $p$-length of $G$ is $\leq 1$ for every prime $p$.

Proof. Suppose $\Phi(G) > 1$. Then there is some prime $p$ and $p$-Sylow subgroup $P$ of $\Phi(G)$ with $P > 1$. Let $Q$ be a $p$-Sylow subgroup of $G$ with $P \leq Q$. Since $Q$ is elementary abelian by assumption, $P$ has a complement in $Q$. By a theorem of Gaschütz [2,I,17.4], $P$ then has a complement in $G$. But this is impossible, since $1 < P \leq \Phi(G)$. Thus $\Phi(G) = 1$. The rest of the statement follows by [2,VI,6.6a]. $\newcommand{\X}{\mathfrak{X}} \newcommand{\Xbar}{\overline{\X}}$$\newcommand{\F}{\mathfrak{F}} \newcommand{\Fbar}{\overline{\F}}$

Lemma 2. Let $G$ be a finite group. Then the following are equivalent:

1. $G \in \X$ (resp. $\Xbar$).

2. If $U \leq G$, then $\Phi(U) = U^\F$ (resp. $\Phi(U) = U^\Fbar$).

$\newcommand{\implies}{\Rightarrow}$ Proof. (1 $\implies$ 2). If $U \leq G \in \X$, then $U \in \X$. Thus all Slow subgroups of $U/\Phi(U)$ have trivial Frattini subgroups, and are elementary abelian. Thus $U^\F \leq \Phi(U)$. By Lemma 1, likewise $\Phi(U) \leq U^\F$. It follows that $\Phi(U) = U^\F$. The proof for $G \in \Xbar$ is analogous.

(2 $\implies$ 1). If $U \leq V \leq G$, then $U^\F \leq V^\F$, as $\F$ is closed under taking subgroups. Thus $\Phi(U) \leq \Phi(G)$; so $G \in \X$. The proof for $G \in \Xbar$ is analogous.

Lemma 3. The class $\X$ is closed under taking quotient groups, and so is $\Xbar$.

Proof. Let $N \lhd G \in \X$, and let $U/N \leq G/N$. By Lemma 2, $\Phi(U) = U^\F$. From this it follows by [1, IV, 1.17] that $(U/N)^\F = (U^FN)/N = \Phi(U)N/N \leq \Phi(U/N)$. Then by Lemma 1, $(U/N)^\F = \Phi(U/N)$. It follows by Lemma 2 that $G/N \in \X$. $\newcommand{\S}{\mathfrak{S}}$

Definition Let $\X' = LF(h)$ be the local class, defined locally by $h$, where $h(p) = \F \cap \S_p$. As $h(p)$ is closed under taking subgroups, $\X'$ is also closed under taking subgroups, by [1,IV,3.14].

Lemma 4. We have $X' = E_\Phi \F$. Thus $E_\Phi \F$ is closed under taking subgroups.

Proof. By [1, IV, (2.9)], $E_\Phi \F$ is a Schunck class. Thus by [1, III, (2.11)] any group $G$ of minimal order in $\X' \setminus E_\Phi \F$ is primitive. If the minimal normal subgroup of $G$ is a $p$-group, then $G/N$ is a $p$-group in $\F$. But then $G \in \F$, contradicting $G \notin E_\Phi \F$. This shows $\X' \subseteq E_\Phi \F$.

[End of second page (p.535)]

Conversely, if $G$ is a group of minimal order in $E_\Phi \F \setminus \X'$, then $G$ is again primitive, so $\Phi(G) = 1$. Thus $G \in \F$. Since it is easy to see that $\F \subseteq \X'$, this again leads to a contradiction. In sum, we conclude that $\X' = E_\Phi \F$.

• Hagen: I hope it is OK with you if others extend this answer into a collaborative translation? I have some time to kill and feel like brushing some of the rust off my German… Nov 3, 2014 at 21:52

I started as follows:

On solvable finite groups, which behave towards the Frattini group like nilpotent groups.

We denote by $\phi(G)$ the Frattini-subgroup of a group $G$. If $G$ is a finite nilpotent group, and if $U\le G$ is a subgroup, $N\le U$ a normal subgroup, it is well known that we have $\phi(U)\le \phi(G)$ and $\phi(G/N)=\phi(G)N/N$. The aim of this note is the characterisation of those finite solvable groups, where the Fratini subgroup also satisfies the above properties.