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The question is as follows:

Suppose group H acts faithfully on the group G via automorphisms and also suppose that $K \vartriangleleft G$ contains $C_G(K)$. And also if we know that $H$ fixes the elements of $K$ i.e. $H K = K$; then how can we show that $H$ is Abelian?

Some definitions:

$H$ acts faithfully on $G$ means that the kernel of the action i.e. $\{ h \in H | hg=g \text{ for all } g \in G \}$ is $\{e\}$, and via automorphism means that for every $x, y \in G$ and $h \in H$ we have $(xy) \cdot h = (x \cdot h)(y \cdot h)$ and each element of $H$ induces an automorphism of $G$. In this case we can suppose that $H$ acts by conjugation and if $K \vartriangleleft G $ then $K$ contains $C_G(K)$ by lemma 7.1 Isaacs "Algebra: A Graduate Course". And $H$ fixes elements of $K$ means that $k\cdot h = k$ for all $k\in K$ and $h\in H$.

Now by knowing all of these how can we prove that $H$ is Abelian?

Thanks!

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    $\begingroup$ I supose that when you say "fix the elements of $\;K\,$", you actually mean $\;K^H\le K\;$, right? Otherwise it must be $\;K=1\;$ as $\;H\;$ acts faithfully on $\;G\;$ by aut. $\endgroup$ – DonAntonio Dec 2 '17 at 23:29
  • $\begingroup$ @DonAntonio Yes sorry, that is what I mean. Now I will correct it. $\endgroup$ – Farrokh Dec 2 '17 at 23:32
  • $\begingroup$ I think you mean $k^h = k$ for all $h \in H$ and $k \in K$. $\endgroup$ – Derek Holt Dec 3 '17 at 8:14
  • $\begingroup$ You have to put some effort into solving the problem yourself,or it will be closed. $\endgroup$ – Derek Holt Dec 3 '17 at 8:24
  • $\begingroup$ Can you give some hint for to prove it? $\endgroup$ – Farrokh Dec 3 '17 at 8:31
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Your claim is not true. You define "$H$ fixes $K$" as $HK=K$ but this is the definition of "$K$ is $H$-invariant" which is not sufficient to show that $H$ is abelian. Consider this example :

$H=Aut(S_3)=S_3$ acts on $G=S_3$ via automorphisms in a faithful fashion. Take $K=G$. Clearly $K$ contains $C_G(K)$ since $K$ is the whole group. Then $HK=K$ but $H$ is not abelian.

Now let's consider the claim with the correct definition of "$H$ fixes $K$", i.e. $k\cdot h = k$ for all $k\in K$ and $h\in H$. In other words, we say "$H$ fixes $K$" if $H$ acts trivially on $K$.

I will prove the claim using three subgroups lemma which is for groups $A,B,C$ if one has $[A,B,C]=1$ and $[C,A,B]=1$ then $[B,C,A]=1$. If it confuses you I can clarify the method and the lemma (for the start, you can check Derek Holt's comment).

First note that the commutator $[K,H,G]=1$ since $[K,H]=1$. We also have $[G,K,H]=1$ since $[G,K]\leq K$ ($K$ is normal) and $[K,H]=1$. Hence we have $[H,G,K]=1$. Since $[H,G]\leq G$ and it commutes with $K$, we get $[H,G]\leq K$ (remember that $K$ contains its centralizer).

We deduce that $[G,H,H]=1$ (since $[G,H]\leq K$). We also have $[H,G,H]=1$ because of the same reason. Hence $[H,H,G]=1$. This means that the subgroup $[H,H]\leq H$ acts trivially on $G$. Since the action is faithful, we deduce that $[H,H]=1$ so $H$ is abelian.

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  • $\begingroup$ Thanks! Today around noon I will let you know the correct definition and condition. $\endgroup$ – Farrokh Dec 5 '17 at 5:07
  • $\begingroup$ Yes your claim is correct! Thanks! I will correct the statement of the question as soon as possible. $\endgroup$ – Farrokh Dec 5 '17 at 12:53
  • $\begingroup$ I answered the question according to the correct statement. $\endgroup$ – Levent Dec 5 '17 at 13:03
  • $\begingroup$ Thanks! Very clever answer! $\endgroup$ – Farrokh Dec 5 '17 at 23:25
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    $\begingroup$ If you think that the answer is helpful and complete, it would be nice to accept it. $\endgroup$ – Levent Dec 6 '17 at 14:19

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