# Group Theory/Characteristic subgroups

Definition (characteristic subgroup):
Let $G$ be a group. A characteristic subgroup of $G$ is a subgroup $H\leq G$ such that $\varphi (H)=H$ for all $\varphi \in \operatorname {Aut} _{\textbf {Grp}}(G)$.
Proposition (characteristic subgroups are normal):
Any characteristic subgroup of a group $G$ is a normal subgroup of $G$.
Proof: This follows since the map $j\mapsto g^{-1}jg$ is a group automorphism of $G$. $\Box$
Definition (characteristically simple):
A group $G$ is called characteristically simple if and only if its only characteristic subgroups are $G$ and $\{e\}$, where $e$ denotes the identity of $G$.
Proposition (characteristically simple groups):
Let $G$ be a characteristically simple finite group, and let $H$ be any of its minimal normal subgroups. Then $G$ is isomorphic to a product of copies of $H$, that is, $G\cong \prod _{\alpha \in A}H$, where $A$ is an index set (of finite cardinality).
Proof: Let $M$ be a subgroup of $G$ maximal subject to the following two conditions:
* $M$ is the direct sum of images of $H$ under a $\varphi \in \operatorname {Aut} _{\textbf {Grp}}(G)$
* $M$ is normal
Suppose that $M\neq G$. Note that the group $\langle \varphi (H)|\varphi \in \operatorname {Aut} _{\textbf {Grp}}(G)\rangle$ is characteristic, so that it equals all of $G$. Hence, we find $\varphi \in \operatorname {Aut} _{\textbf {Grp}}(G)$ such that $\varphi (H)$ is not a subgroup of $M$. Since $\varphi$ is an automorphism, $\varphi (H)\triangleleft G$, so that $M\cap \varphi (H)=\{e\}$. Since the product of normal subgroups is normal, we conclude that the product subgroup $M\varphi (H)\triangleleft G$ is a normal subgroup that is a direct product of homomorphic images of $H$ in $G$, in contradiction to the maximality of $M$ with these properties. Hence, $M=G$ and we are done. $\Box$
Proposition (minimal normal subgroups of a characteristically simple groups are simple):
Let $G$ be a characteristically simple group, and let $H\triangleleft G$ be a minimal normal subgroup of $G$. Then $H$ is simple.
Proof: $\Box$
Proposition (powers of characteristically simple groups are characteristically simple):
We conclude:
Theorem (structure theorem of finite, characteristically simple groups):
The finite, characteristically simple groups are precisely the powers of simple groups.
Proof: We have seen that each characteristically simple finite group is the direct product of copies of isomorphic images of any of its minimal normal subgroups, and that the latter are always simple in characteristically simple groups. We conclude that each finite, characteristically simple group is a power of simple groups. Conversely, let $H$ be a simple group, $n\in \mathbb {N}$, and set
    $G:=H^{n}:=\overbrace {H\times \cdots \times H} ^{n{\text{ times}}}$. $\Box$

## Exercises

* Prove that all subgroups of $\mathbb {Z} _{6}$ are characteristic.
* Let $H,L$ be two finite simple groups such that $|L|$ is divisible by a prime number $p$ that does not divide $|H|$. Use the structure theorem for characteristically simple groups to prove that $H\times L$ is not characteristically simple.
* Prove that a subgroup of a characteristically simple group need not be characteristically simple.
* Prove that the product of characteristically simple subgroups whose minimal normal subgroups are not isomorphic is not characteristically simple.