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Here, I wanted to verify that:

The property of being characteristic is a transitive relation among subgroups of a group $G$.

For subgroups $N,H\leq G$, we have $N$ char $G$ and $H$ char $N$. So $N$ char $G$ implies: $$\forall\psi\in Aut(G); \psi(N)=N$$ $H$ char $N$, so for all elements in $Aut(N)$, $H$ is remained never changing. Especially, when I take $\psi'=\psi|_{N}$ then $\psi':N\to Aut(N)$ would be in $Aut(N)$ and $\psi'(H)=H$. Since the last equality is true for $H$ and the maps, caused by restriction on $N$, then I have $H$ char $G$.

Honestly, I am inly not satisfied form the conclusion here and think I am losing something. Thanks for your hints.

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up vote 6 down vote accepted

So we have $\,H\,$ char $\,N\,$ char $\,G\,$ . Let

$$\phi\in \operatorname {Aut}(G)\Longrightarrow \phi(N)=N\,\,,\,\text{since}\,\,N\,\,\text{is characteristic in}\,\,G$$

but this means $\,\left.\phi\right|_N\in\operatorname{Aut}(N)\,$ , so

$$\phi(H)=\left.\phi\right|_N(H)\stackrel{\text{since}\, H\,\mathbf{char}\, N}=H\Longrightarrow H\,\,\mathbf {char}\,G$$

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Thank you for the time andmaking me sure. – S. Snape Sep 24 '12 at 9:25

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