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Let $ A \trianglelefteq G $ and $ B \trianglelefteq A $ a Sylow normal subgroup of $ A $. My textbook says then that $ B \trianglelefteq G $.

I don’t understand why that is.

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Since $B$ is a normal sylow $p$-subgroup there is one $p$-sylow subgroup in $A$ since all the sylow $p$-subgroups are conjugated. Let $g\in G, gBg^{-1}\subset A$ since $A$ is normal in $G$ and is a $p$-sylow subgroup since it has the same cardinal than $B$ thus is $B$ since $B$ is the unique $p$-sylow subgroup of $A$,

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A subgroup $H$ of a group $G$ is said to be a characteristic subgroup if it is fixed under all automorphisms of $G$ (not just conjugation). Precisely, if $\sigma \in Aut(G)$, then $\sigma H = H$ as a set.

Characteristic subgroups are of course normal, because conjugation is an automorphism.

The point of this definition is that, although the condition of being a normal subgroup is not transitive, the condition of being characteristic is. Why?

Suppose that $B$ is a normal subgroup of $A$ and $C$ is a normal subgroup of $B$. Let's try to prove that $C$ is normal in $A$ and see what fails: let $a \in A$, and consider that $aBa^{-1} = B$. But now the conjugation action of $a$ on $C$ is not necessarily induced by conjugating $C$ with an element of $B$. However, $b \to aba^{-1}$ is still an automorphism of $B$. So if $C$ was characteristic in $B$, then it would still be fixed by conjugation by $b$.

Thus it follows that $aCa^{-1} = C$, for all $a \in A$, under the condition that $C$ is not just normal in $B$, but characteristic. So we have proven:

Lemma: Let $B$ be a normal subgroup of $A$, and let $C$ be a characteristic subgroup of $B$. Then $C$ is normal in $A$.

The key philosophy now is that canonically defined subgroups are characteristic. (For example, the center of a group, the commutator subgroup, or the socle of a group.)

In particular, when there is a normal $p$-Sylow subgroup, it is the unique $p$-Sylow subgroup. Thus it is characteristic, and you can apply the lemma.

For more see here: https://en.wikipedia.org/wiki/Characteristic_subgroup

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  • $\begingroup$ I think you meant to write that characteristic subgroups are normal, not the other way around (in the second paragraph). $\endgroup$
    – kccu
    Dec 28, 2015 at 4:52
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let $x\in G$ we'll show that $xBx^{-1}=B$, we have :

$B$ subgrouf of $A$ means $xBx^{-1}\subset xAx^{-1}=A$ ( because $A \trianglelefteq G$, so $xBx^{-1}\subset A$, which means that $B$ and $xBx^{-1}$ are two $p$-subgroups of sylow of $A$. So $B=xBx^{-1}$ because $B \trianglelefteq A$.

Finally $B \trianglelefteq G$.

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