Let's take a group $G$. We say that the subgroup $H \leq G $ is fully characteristic if $$\forall \phi \in \mathrm{End} (G) : \phi(H) \subseteq H.$$

Is this the fully characteristic subgroup normal?

A first thought is to apply the Theorem :

$$\forall g \in G, \forall h \in H :ghg^{-1}\in H. $$ But how could this help as? I have definitely stuck.

Update: Could we find a normal subgroup $H\trianglelefteq G$ such that $H$ is not strictly characteristic? (so the reverse statement is not valid)

Thank you.

P.S.: I apologize for not having any other progress, but I don't know how to continue.

  • 1
    $\begingroup$ Hint: Think of inner automorphisms. $\endgroup$
    – asdq
    May 14 '18 at 21:21

Hint: the map $\phi_g\colon G\to G$, $\phi_g(x)=gxg^{-1}$ is an endomorphism of $G$ (actually an automorphism).

  • $\begingroup$ Thank you for your answer. For all $g \in G, \forall h\in H: \phi _G (h) = ghg^{-1} \in \phi (H) \subseteq H \implies ghg^{-1}\in H. $ Right? $\endgroup$
    – Chris
    May 14 '18 at 21:43
  • 1
    $\begingroup$ @Chris If you prove that $\phi_g$ is an endomorphism, you're done with the argument you showed. $\endgroup$
    – egreg
    May 14 '18 at 21:46

Consider these:

  • A subgroup is fully characteristic iff it is invariant under all endomorphisms

  • A subgroup is normal iff it is invariant under all inner automorphisms

  • Every inner automorphism is an endomorphism

  • 1
    $\begingroup$ The main catch to consider is that "invariant" is generally used in a stronger sense for automorphisms (requiring $\varphi(H) = H$ rather than just $\varphi(H)\subseteq H$). But once we put "for all" there, it works out anyway. $\endgroup$ May 15 '18 at 6:55

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