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The proposition that I want to prove is the following: If $G$ is a group and $x\in G$ is such that $C_G(x)$ is a normal subgroup, then show that there's an abelian normal subgroup of $G$ containing $x$.

First of all I've thought that the subgroup we're looking for is also a subgroup of $C_G(x) $.

I've thought that $C_G(x)=C_G(S)$ where $S=\langle x\rangle$, i.e. $S$ is the cyclic subgroup generated by $x$, and then $G/C_G(S)$ is isomorphic (by the N/C Theorem) to a subgroup of $Aut(S)$, and $Aut(S)$ is abelian because $S$ is cyclic. That means that $G' \leq C_G(S)$, where $G'$ is the derived group of G, hence every commutator of $G$ commutes with $x$. However the derived group might be a normal subgroup contained in the centralizer of $x$, but it doesn't necessarily contain $x$ and isn't necessarily abelian.

$S$ on the other hand is abelian, contains $x$ and is inside its centralizer, but I can't see why it should be normal.

Thank you in advance for any help.

Additional inquiry: What about the inverse of this proposition? I can't really prove it and I can't think of a counter-example either.

Nothing just that
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    Use \langle and \rangle, not < and > – Arturo Magidin Dec 02 '20 at 00:12
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    Hint: try the center of $C_G(x)$. – Jason DeVito - on hiatus Dec 02 '20 at 00:18
  • @JasonDeVito: By your hint, I derived that if $z\in C_G(C_G(x))$ then $zx$ and $xz$ and $xzx^{-1}$ all are also in $C_G(C_G(x))$. You meant that this group is the wanted normal subgroup? – C.F.G Dec 02 '20 at 09:43
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    @C.F.G.: Well, I meant $Z(C_G(x))$, but yes, this is the desired normal subgroup. – Jason DeVito - on hiatus Dec 02 '20 at 12:35
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    Use that if $X \leq Y \leq Z$ are groups, $X$ is characteristic in $Y$, and $Y$ is normal in $Z$, then $X$ is normal in $Z$. In addition, see here for more info https://math.stackexchange.com/questions/816442/finite-groups-of-which-the-centralizer-of-each-element-is-normal. In stead of $Z(C_G(x))$ you could take $\langle Cl_G(x) \rangle$, the group generated by all the conjugates of $x$. In $Cl_G(x)$ every two elements commute, and it is a normal set. – Nicky Hekster Dec 02 '20 at 13:21
  • Thanks for the help, I should have probably considered the center myself, it's one of the few choices for abelian subgroups. – Nothing just that Dec 02 '20 at 16:57
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    Instead of editing in an additional question, please ask it in a separate post. – Shaun Dec 02 '20 at 17:16

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