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Let $G$ be a finite group, $N$ be a minimal normal subgroup of $G$, and $M$ be a smallest normal subgroup of $G$. What's the difference between a smallest normal subgroup and a minimal normal subgroup? What's the definition of a smallest normal subgroup?

This definition needed for this 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 $4$.

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

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    $\begingroup$ There are two possible interpretations that I can think of. 1) Minimal means that there is no normal subgroup of $G$ that is a proper subgroup of $N$, while smallest means that any other normal subgroup of $G$ has $M$ as a subgroup. 2) Minimal means that $N$ doesn't have any non-trivial normal subgroups in its own right, while smallest means that $M$ is a normal subgroup of any normal subgroup of $G$. Whichever interpretation is used, the main message is that (some of) the subgroups of $G$ are arranged in a poset, and minimal just means that none are smaller, while smallest means smallest. $\endgroup$ – Arthur Aug 19 '15 at 10:35
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    $\begingroup$ So according to the second interpretation $N$ is a simple group. $\endgroup$ – Stefan Hamcke Aug 19 '15 at 10:40
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    $\begingroup$ It is not possible to answer questions like this unless you provide more context, by quoting the complete sentence in which the phrase occurs. $\endgroup$ – Derek Holt Aug 19 '15 at 11:27
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Smallest normal subgroup $N$: It means that if $H \trianglelefteq G$ then $N \leq H$. Note that the intersection of two normal subgroups is normal, so $N = \bigcap_{H \trianglelefteq G} H$. You usually want the smallest normal subgroup having some specific property. It is possible that $N = \{e\}$.

Minimal normal subgroup $M$: there are no nontrivial normal subgroups of $G$ properly contained in $M$. There can be normal subgroups that are smaller than $M$ (in terms of order). There can also be different, nonisomorphic minimal normal subgroups.

For example, consider $\mathbb{Z}_{2} \times S_{5}$. We have that $\{e\}\times A_{5}$ and $\mathbb{Z}_{2} \times \{e\}$ are both minimal normal subgroups of $G$. Along with $\mathbb{Z}_{2} \times A_{5}$, these are all the nontrivial normal subgroups of $G$. The smallest normal subgroup is the intersection of these three subgroups, which is $\{e\}\times\{e\}$.

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  • $\begingroup$ This definition neede for this 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 $. Proof: Let $ K/L $ be a chief factor of $ G $. We proceed by induction on $ |K| $. Let $ M $ be the smallest normal subgroup of $ K $ such that $ K/M $ is nilpotent, and so $ M \unlhd G $ and $ M \leq L $. $\endgroup$ – Soroush Aug 19 '15 at 10:58
  • $\begingroup$ What's the mean $ G $-isomorphis ? $\endgroup$ – Soroush Aug 19 '15 at 11:30

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