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$H$ 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

Shaun
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MathMan
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  • you should start with Suppose $h\in H$ and $g\in G$.. :D –  Jun 27 '14 at 09:46
  • You're correct, normality is not transitive, so $H\triangleleft K\triangleleft G$ does not imply $H\triangleleft G$. You're trying to show that $H$ is closed under $G$-conjugations, which are $K$-automorphisms since $K\triangleleft G$. Note this question probably has many duplicates by now. – anon Jun 27 '14 at 09:49
  • I have a query. $H$ is characteristic subgroup of $K \implies kHk^{-1} = H\forallk \in K$ and not $\forall g \in G$? Is this right? – MathMan Jun 27 '14 at 09:57
  • @VHP : yes.... Not for all $g\in G$ –  Jun 27 '14 at 09:58
  • But you wrote : As $H$ is characteristic subgroup of $K$ we see that $\psi(H)=H$ i.e., $gHg^{-1}=H \forallg \in G$. Shouldn't it be instead like :

    $H$ is characteristic subgroup of $K \implies kHk^{-1} = H\forall k \in K$?

     – MathMan Jun 27 '14 at 10:05
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    @VHP Remember that $\psi$ is the map $K\rightarrow K, n\mapsto gng^{-1}$, which defines an automorphism of $K$. – user1729 Jun 27 '14 at 10:10
  • Oh ok, i think I got confused with the definition of $Inn (K)$ where $K$ is a subgroup of $G$

    $Inn(K) = gkg^{-1}, ~\forall~k \in K, g \in G$. Correct now?

     – MathMan Jun 27 '14 at 10:15
  • No No.. please note that inner automorphisms are functions where as you have written $gkg^{-1}$.. –  Jun 27 '14 at 10:30
  • I am sorry, I don't understand. Can you elaborate a bit please? – MathMan Jun 27 '14 at 10:32
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    You might want to have a look at http://math.stackexchange.com/questions/689555/prove-that-h-is-a-normal-subgroup-of-g/689772#689772 – Nicky Hekster Jun 27 '14 at 10:47
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    This is a FAQ. http://math.stackexchange.com/questions/647058/ http://math.stackexchange.com/questions/800164/ – Jack Schmidt Jun 27 '14 at 13:54

4 Answers4

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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, $H\unlhd G$.

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

    $H$ is characteristic subgroup of $K \implies kHk^{-1} = H$ and not $\forall g \in G$? Am i confused?

     – MathMan Jun 27 '14 at 09:55
  • You are confused i guess... do you see that $\psi$ is an automorphism of $K$ –  Jun 27 '14 at 09:58
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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 $8$.)

user1729
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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 $G$.

Hans
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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$.

sez
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