Prove that if $H$ is a characteristic subgroup of $K$, and $K$ is a normal subgroup of $G$, then $H$ is a normal subgroup of $G$

Question

$H$ is a characteristic subgroup of $K$ if $\Phi(H)=H~\forall~\Phi\in Aut(K).$ Prove that if $H$ is a characteristic subgroup of $K$, and $K$ is a normal subgroup of $G$, then $H$ is a normal subgroup of $G$

Attempt: Suppose $h \in H, k \in K$

Given that $\Phi(H) = H ~~\forall~~ \Phi~\in Aut(K).$

Since, $Inn (K) \subset Aut(K) \implies k^{-1}Hk = H \implies H \vartriangleleft K$ .... $(1)$

Also, it's given that $K \vartriangleleft G$ .....$(2)$

From $(1),(2): H \vartriangleleft K \vartriangleleft G$

But, I don't think this means that $H \vartriangleleft G?$ Is there something which I am missing?

Thank you for your help

Answer 1

For $g\in G$ Consider $\psi :K\rightarrow K$ with $n\rightarrow gng^{-1}$.

This is well defined (???)and an automorphism of $K$(???).

As $H$ is characteristic subgroup of $K$ we see that $\psi(H)=H$ i.e., $gHg^{-1}=H$.

As $g\in G$ is arbitrary, we see that $gHg^{-1}=H$ for all $g\in G$. Thus, $H\unlhd G$.

Answer 2

Hint: If $g\in G$ and $K$ is a normal subgroup of $G$ then the conjugation action of $g$ induces an automorphism of $K$.

(Also, having a chain $H\lhd K\lhd G$ does not imply that $H\lhd G$. I believe that the standard example is the semidirect product $(H\times H)\rtimes \mathbb{Z}_2$, where the action of $\mathbb{Z}_2$ swaps the copies of $H$. Taking $H$ to be cyclic of order two gives a counter-example of order $8$.)

Answer 3

For any $g \in G$ the map $x \mapsto g^{-1} x g$ is an automorphism of $G$. Since $K$ is normal, it is fixed under this automorphisms, i.e. they are automorphisms of $K$. Thus $H$ is fixed under those automorphisms, which means that $H$ is normal in $G$.

Answer 4

Let $g$$\in$$G$. Then conjugation by $g$ maps $K$ onto itself as it is normal and the restriction of this conjugation map to $K$ is an automorphism of $K$, not necessarily an inner automorphism of $K$. Since $H$$char$$K$, $H$ is mapped onto itself by this automorphism of $K$ for all $g$$\in$$G$ and hence $H$$^{g}$=$H$ implying that $H$$\triangleleft$$G$.