TITLE: 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 [5 upvotes]: $H$ 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

REPLY [2 votes]: 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$.