theorem: Let $ G $ be solvable with $ \Phi(G)=1 $ and assume that each minimal normal subgroup has prime order or order $ 4 $. Then every chief factor of $ G $ has prime order or is $ G $-isomorphic to a minimal normal subgroup of $ G $ of order $ 4 $.

For proof: Let $ K/L $ be a chief factor of $ G $. We proceed by induction on $ \vert K \vert $. Let $ M $ be the smallest normal subgroup of $ K $ such that $ K/M $ is nilpotent, and so $ M \lhd G $ and $ M \leq L $.

What's mean smallest normal subgroup and $ G $-isomorphis ?


The smallest normal subgroup $M$ of $K$ such that $K/M$ is nilpotent is the intersection of all normal subgroups $L$ of $K$ such that $K/L$ is nilpotent. If $K$ is finite then it can be proved that $K/M$ really is nilpotent. (This is not necessarily true if $K$ is infinite.) To prove it in the finite case it is sufficient to prove that if $K/L_1$ and $K/L_2$ are nilpotent then so is $K/(L_1 \cap L_2)$.

Two chief factors of $G$ are $G$-isomorphic if there is an isomorphism $f$ between them such that $f(n^g)=f(n)^g$ for all $g \in G$.

  • $\begingroup$ This is primary $ M \leq L $, but why $ M \lhd G $ ? $\endgroup$ – Soroush Aug 19 '15 at 18:58
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    $\begingroup$ $M$ is characteristic in $K$ and $K$ is normal in $G$, so $M$ is normal in $G$. $\endgroup$ – Derek Holt Aug 20 '15 at 19:30
  • $\begingroup$ Why $ M $ is characteristic in $ K $ ? $\endgroup$ – Soroush Aug 21 '15 at 15:32
  • $\begingroup$ That's obvious from the definition of $M$. $\endgroup$ – Derek Holt Aug 21 '15 at 20:19
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    $\begingroup$ Yes, sorry replace $M/K$ by $K/M$. $\endgroup$ – Derek Holt Sep 18 '15 at 8:57

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