# Smallest normal subgroup and minimal normal subgroup, what's the difference?

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 $$M \leq L$$.

• 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. – Arthur Aug 19 '15 at 10:35
• So according to the second interpretation $N$ is a simple group. – Stefan Hamcke Aug 19 '15 at 10:40
• It is not possible to answer questions like this unless you provide more context, by quoting the complete sentence in which the phrase occurs. – Derek Holt Aug 19 '15 at 11:27

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 $$\{e\}\times\{e\}$$.
• 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 $M \leq L$. – Soroush Aug 19 '15 at 10:58
• What's the mean $G$-isomorphis ? – Soroush Aug 19 '15 at 11:30