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Let $G$ be a group and $H$ a subgroup of $G$ .Prove that $H$ is not a subgroup of $C_G(H)$ where $C_G(H)=\{g \in G|g^{-1}hg=h, \forall h \in H \}$

Assertion: If $H \leqslant C_G(H)$ then $H$ is abelian

Proof:

Suppose that $H \leqslant C_G(H)$.

$\forall h \in H $ we have that $h \in C_G(H)$ thus $h^{-1}h_1h=h_1$,$\forall h_1 \in H$ $\Longrightarrow h_1h=hh_1 \forall h_1,h \in H$ ,thus $H$ is abelian.

If $H$ is not abelian then $H$ is not a subgroup of $C_G(H)$

Now we can take $SL_2( \mathbb{Z}) \leqslant Gl_2(\mathbb{Z})$ as a counterexmple.

Is this approach valid or do i have to come up with another approach?

Thank you in advance!

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    Can you please state *exactly* what you are supposed to prove? Sometimes subgroups are contained in their centralizers (when they are Abelian, as you have pointed out), other times not. – 2017-02-19
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    i stated already in the question what i have to prove but i proved that the property that H is subgroup of $C_G(H)$ is valid if H is abelian thus its easier for me then to find an appropriate counterexample . – 2017-02-19
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    How can you be asked to prove that "$H$ is not a subgroup of $C_G(H)$," when sometimes it is, sometimes it isn't, depending on $G$ and $H$? I'm suggesting that there is a missing phrase ("Prove *or disprove*, that...), or condition ("...if and only if $H$ is not Abelian)" that you haven't included. – 2017-02-19
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    i proved that if H is a subgroup of CG(H) then H is abelian.I can edit my answer and add an assertion. – 2017-02-19
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    @pjs36 I expect it was a badly worded exercise. – 2017-02-19
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    @capo: This question makes no sense: in the first paragraph you want to prove that $H$ is not a subgroup of $C_G(H)$; in the second, you suppose that $H$ is a subgroup of $C_G(H)$. Do *you* understand what you are asking? – 2017-02-24
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    i just proved that this property is true for abelian groups.its for my own help to find a counterexample' – 2017-02-24

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