# Let $M$ be the smallest normal subgroup of $K$ such that $K/M$ is nilpotent. What's mean smallest normal subgroup?

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