Show that if \(H \leq G\), \(K\normal G\) is a normal subgroup, and \(H~\text{char}~K\) then \(H\) is normal in \(G\).
So normality is not transitive, but strengthening one to “characteristic” gives a weak form of transitivity.
Solution
Assume \(H\operatorname{char}K\) and \(K\trianglelefteq G\).
1. For every \(g\in G\), conjugation by \(g\) restricts to an automorphism of \(K\).
Because \(K\trianglelefteq G\), \[ gKg^{-1}=K. \] Thus the inner automorphism \[ c_g:G\to G,\qquad x\mapsto gxg^{-1} \] restricts to an automorphism \(c_g|_K\in\operatorname{Aut}(K)\).
2. Every such conjugation preserves \(H\).
Since \(H\) is characteristic in \(K\), every automorphism of \(K\) maps \(H\) to itself. Applying this to \(c_g|_K\) from 1 gives \[ gHg^{-1}=H \] for every \(g\in G\).
3. Therefore \(H\trianglelefteq G\).
The equality \(gHg^{-1}=H\) for all \(g\in G\) is exactly the definition of normality.