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If $A$ is a non-abelian normal simple subgroup of a group $G$ and $C$ is the centralizer of $A$ in $G$, could someone explain why $A\cap C = 1$? I'm basically trying to understand this answer.

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    The intersection of a subgroup $H$ of a group $G$ and its centralizer in $G$ is just the center of the group $H$. Now, use the fact that non-abelian simple groups are centerless. – Geoffrey Trang May 12 '20 at 14:12
  • @GeoffreyTrang Understood, thanks! The only subgroups of a simple group $H$ are the trivial group and the whole group $H$. However, the center of a group $Z(H)$, which is a subgroup of $G$, is necessarily abelian, so the trivial group is the only choice. –  May 12 '20 at 14:30
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    It isd simply not true that the only subgroups of a nonabelian simple group $H$ are the trivial group and the whole group $H$. Those are its only normal subgroups. – Derek Holt May 12 '20 at 18:45

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In general, if $H$ is a subgroup of $G$, then $Z(H)=H \cap C_G(H)$. So if $H$ is non-abelian simple, we must have $Z(H)=1$. Normality of $H$ is not needed.

Nicky Hekster
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    Thanks for pointing this out! Yes, indeed, normality isn't a necessity here. I had missed that. :) –  May 12 '20 at 14:40
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The centraliser $C$ of a subgroup $A$ is a normal subgroup of its normaliser, which is $G$ if $A$ is normal in $G$. As $A$ is non-abelian, $C\cap A\ne A$, and it is a normal subgroup of $A$, so $C\cap A=\{1\}$ if $A$ is simple.

(My thanks to Arturo Magidin for pointing a wrong argument in the initial version of this answer).

Bernard
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  • Typo: “which is $A$ if $A$ is normal in $G$”; the first $A$ should be a $G$. But it is not true that $C$ must be trivial... it is the intersection with $A$ that must be trivial, as $C$ does not contain $A$. – Arturo Magidin May 12 '20 at 17:19
  • @ArturoMagidin: What do you mean with ‘the first $A$ should be a $G$’? – Bernard May 12 '20 at 17:55
  • “ of its normalizer, which is $A$ if $A$ is normal in $G$.” If $A$ is normal in $G$, then its normalizer is $G$, not $A$. – Arturo Magidin May 12 '20 at 18:01
  • Oops! Yes, I didn't check my shortcut; Thanks! I'll fix it. – Bernard May 12 '20 at 18:07
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Claim. $A \cap C \lhd A$.
Proof. Let $a \in A$ be arbitrary. Let $g \in A \cap C$.
In particuar, $g \in C$ and thus, $aga^{-1} = g \in A \cap C$. (Since $a$ and $g$ commute.)

As $A$ is simple, this forces $A \cap C$ to be either $(1)$ or $A$. We show that the latter is not possible and thus, prove the result.

Suppose $A \cap C = A$, then we would have that $A \le C$.
Claim. $A$ is abelian. (This is the contradiction.)
Proof. Let $a_1, a_2 \in A$. Then, $a_2 \in C$ as well.
Since $C$ is the centraliser of $A$, we see that $a_1 a_2 = a_2 a_1$.
As $a_1, a_2$ were arbitrary, this proves the claim.

As we have arrived at a contradiction, we are done.

Aryaman Maithani
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$A\cap C\triangleleft A\stackrel{\text{A simple}}\implies A\cap C=\{1\}$.

Note: we can't have $A\cap C=A$ because $A\cap C$ is abelian.