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I would appreciate some help in translating the statements of the theorems contained in the following paper of Gaschütz on the Frattini subgroup: "Über die $\Phi$-Untergruppe endlicher Gruppen".

Many thanks and I'm willing to offer compensation for this.

Satz 1 Satz 2 Satz 3

Satz 4 Satz 5 Satz 6 Satz 7 Satz 8 Satz 9 Satz 10 Satz 11 Satz 12 Satz 13 Satz 14 Satz 15 Satz 16 Satz 17

the_fox
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1 Answers1

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Feel free to edit , as I am not about the translation of certain terms and I cannot read german fractura very well. (Perhaps you might also want to leave a comment for saying what you changed and why you changed it.)

I denote the words I was not sure how to translate and comments with questionmarks and $[]$ brackets.

Definitions:

  • $A(G)$: the automorphism group of $G$.
  • $J(G)$: the group of inner automorphisms of $G$.
  • $Z(G)$: the centre of $G$.
  • $K(G)$: the commutator subgroup of $G$.
  • $L(G)$: the maximal nilpotent subgroup of $G$.
  • $\Phi(G)$: the intersection of the maximal subgroups of $G$.
  • $\Delta(G)$: the intersection of the maximal and non-invariant subgroups of $G$.
  • If $N$ is a normal subgroup of $G$, and $G=N\overline{G}$, where $\overline{G}$ is a subgroup of $G$ and $\overline{G}\subset G$ ($N\cap\overline{G}=1$, resp.), we say that $G$ splits (decomposes, resp.) over $G$.
  • If $\Gamma$ is a subgroup of $A(G)$, a group $R$ is $\Gamma$-fully-reducible if $R$ is a product of minimal subgroups of $R$ invariant under $\Gamma$. In particular, if $R$ is a normal subgroup of a group $G$, and $\Gamma$ the group of automorphisms of $R$ induced by transformations by the elements of $G$ [I take this to mean conjugation. -BMS], and $R$ is $\Gamma$-fully-reducible, we say that $R$ is $G$-fully-reducible.
  • Let $N$ be an arbitrary normal subgroup of $G$. $S_{G}(N)$ denotes the product of the minimal normal subgroups of $G$ in $N$. $S_{G}(N)$ is the maximal $G$-fully-reducible normal subgroup contained in $N$. We denote by $R_{G}(N)$ the intersection of the normal subgroups of $G$ maximal in $N$. $N/R_{G}(N)$ is $G/R_{G}(N)$-fully-reducible, and $R_{G}(N)$ contains no normal subgroup $M$ of $G$ with $G/M$-fully-reducible factor group $N/M$. We call $S_{G}(G)$ the Sockel of $G$ and write simply $S(G)$.

Theorem 1: Let $N$ be a normal subgroup of $G$. Then $N \subseteq \Phi(G)$ if and only if $G$ does not split over $N$.

Theorem 2: Let $\Psi$ be a normal subgroup of $G$, $\Psi\subseteq \Phi(G)$. Then $\Phi(G/\Psi) = \Phi(G) /\Psi$.

Theorem 3: If $\sigma$ is a homomorphism with domain $G$, then $$(2.1) \qquad\qquad\qquad\qquad\sigma\Phi(G)\subseteq \Phi(\sigma G)\;.$$

Theorem 4: $K(G) \cap Z(G) \subseteq \Phi(G)$.

Theorem 5: If $N$ is a normal supgroup and $U$ a subgroup of $G$, and $N \subseteq \Phi(U)$, then $N \subseteq \Phi(G)$.

Theorem 6). If $G = M \times N$. then $\Phi(G) = \Phi(M) \times \Phi(N)$.

Theorem 7: If $A$ is an Abelian normal subgroup of $G$, and $$(2.5)\qquad\qquad\qquad A \cap \Phi(G) = 1\;,$$ then $G$ decomposes over $A$. Moreover, $A$ is $G$-fully-reducible.

Theorem 8: If $A$ is an Abelian normal subgroup of $G$, then $R_{G}(A) \subseteq \Phi(G)$.

Theorem 9: If $L$ is nilpotent, then $\Phi(L) = 1$ if and only if $L$ is a direct product of groups of prime order.

Theorem 10: Let $M$ and $\Psi$ be normal supgroups of $G$, $\Psi \subseteq M \cap \Phi(G)$ and $M / \Psi$ nilpotent. Then $M$ is nilpotent.

Theorem 11: Let $\Psi$ be a normal supgroup of $G$, $\Psi \subseteq \Phi(G)$. Then $$(4.1)\qquad\qquad\qquad J(\Psi) \subseteq \Phi(A(\Psi))\;.$$

Theorem 12: In a $\Phi$-free group $F$, $L(F)$ is the product of the minimal Abelian normal subgroups of $F$.

Theorem 13: $L(G) / \Phi(G)$ is the product of the minimal Abelian normal subgroups of $G/\Phi(G)$.

Theorem 14: A group is $\Phi$-free if and only if it decomposes over the Abelian component of its Sockel.

Theorem 15: In a $\Phi$-free group $F$,

$$(5.4) \qquad\qquad\qquad Z(F/Z(F)) = Z(F) / Z (F)$$ and

$$(5.5) \qquad \qquad\qquad Z (F) = \Delta(F)\;.$$

Theorem 16: For an arbitrary $G$, $\Delta(G)$ is nilpotent.

Theorem 17: Let $S$ be a direct product of simple groups and $\{\Gamma\}$ a complete system of representatives of the classes of $A(S)$ conjugated subgroups of $A(S)$ with the following properties:

$(I): \quad S_\alpha$ is $\Gamma_\alpha$-fully-reducible

$(II): \quad J(S_\eta) \subseteq \Gamma$

If $\Gamma$ runs through this system of representatives, then the groups $F_\Gamma$ with the elements $(a,\gamma)$, $a\in S_\alpha$, $\gamma \in \Gamma$, and the operation defined by

$$(a,\gamma) (b,\delta) = (ab^{\gamma_\alpha},\gamma \delta)$$

are a complete system of representatives of the types of the $\Phi$-free groups of a Sockel of type $S$.

the_fox
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flawr
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    Your $\mathcal{R}$ in Thm. 1 should be $\mathcal{N}$. The paper contains a definition of splits over $\mathcal{N}$: If $\mathcal{N}$ is a normal subgroup of $\mathcal{G}$, and $\mathcal{G}=\mathcal{N}\overline{\mathcal{G}}$, where $\overline{\mathcal{G}}$ is a subgroup of $\mathcal{G}$ and $\overline{\mathcal{G}}\subset\mathcal{G}$ ($\mathcal{N}\cap\overline{\mathcal{G}}=1$, resp.), we say that $\mathcal{G}$ splits (decomposes, resp.) over $\mathscr{N}$. I’ve no idea what the proper English terminology is; this is very much not my field. – Brian M. Scott Nov 17 '14 at 17:10
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    In Theorem 3: homomorphic. In theorem 4, the letters are $\mathfrak{K}$, $\mathfrak{Z}$, $K$ and $Z$. – Daniel Fischer Nov 17 '14 at 17:10
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    In my other comment I used split for zerspalten and decompose for zerfallen. The latter appears in Thm. 7. – Brian M. Scott Nov 17 '14 at 17:14
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    $A(\mathcal{G})$ is the automorphism group of $\mathcal{G}$. If $\Gamma$ is a subgroup of $A(\mathcal{G})$, a group $\mathcal{R}$ is $\Gamma$-fully-reducible if $\mathcal{R}$ is a product of minimal subgroups of $\mathcal{R}$ invariant under $\Gamma$. In particular, if $\mathcal{R}$ is a normal subgroup of a group $\mathcal{G}$, and $\Gamma$ the group of automorphisms of $\mathcal{R}$ induced by transformations with the elements of $\mathcal{G}$ [I take this to mean conjugation], and $\mathcal{R}$ is $\Gamma$-fully-reducible, we say that $\mathcal{R}$ is $\mathcal{G}$-fully-reducible. – Brian M. Scott Nov 17 '14 at 17:23
  • Feel free to edit now, I incorporated your suggestions where I knew it, but I hope you can contribute where I am still not sure. – flawr Nov 17 '14 at 17:29
  • I'll give it a shot at writing the theorems as they would appear in a modern textbook. – the_fox Nov 17 '14 at 17:30
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    I think that I’ve done what I can now. I left Sockel untranslated but added the definition in the paper. – Brian M. Scott Nov 17 '14 at 18:11
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    There's not much point in retaining the original text's fraktur, or the changes in letters in every theorem. I wish I'd be allowed to make the changes I want so that it reads easily. – the_fox Nov 17 '14 at 18:17
  • @the_fox: There’s no fraktur in the answer as it now stands. – Brian M. Scott Nov 17 '14 at 18:19
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    There's mathcal, which I find completely unnecessary. Also, the operator names are strange. – the_fox Nov 17 '14 at 18:20
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    @the_fox: I can get rid of the mathcal easily enough; it wasn’t my idea anyway. Hang on a bit. – Brian M. Scott Nov 17 '14 at 18:22