Normalizer of Sylow p-Subgroup

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Theorem

Let $P$ be a Sylow $p$-subgroup of a finite group $G$.

Let $\map {N_G} P$ be the normalizer of $P$.


Then any $p$-subgroup of $\map {N_G} P$ is contained in $P$.


In particular, $P$ is the unique Sylow $p$-subgroup of $\map {N_G} P$.


Proof

Let $Q$ be a $p$-subgroup of $N = \map {N_G} P$.

Let $\order Q = p^m, \order P = p^n$.

By Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:

$P \lhd \map {N_G} P$

thus by Subset Product with Normal Subgroup as Generator:

$\gen {P, Q} = P Q$

Thus by Order of Subgroup Product:

$P Q \le G: \order {P Q} = p^{n + m - s}$

where $\order {P \cap Q} = p^s$.

Since $n$ is the highest power of $p$ dividing $\order G$, this is possible only when $m \le s$.

Since $P \cap Q \le Q, s \le m$ thus we conclude that $m = s$ and therefore $P \cap Q = Q$.

Thus from Intersection with Subset is Subset‎:

$Q \subseteq P$


In particular, if $Q$ is a Sylow $p$-subgroup of $\map {N_G} P$, then $Q = P$.

$\blacksquare$


Sources

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