Exercise 4.5.50 (Conjugate normal subsets of a $p$-Sylow)

Proof. Suppose that if $U$ and $W$ are normal subsets of a Sylow $p$-subgroup $P$ of $G$. By definition,

If $U$ is conjugate to $W$ in $N_G(P)$, there is some $g\in N_G(P)$ such that $\mathit{𝑔𝑈}{g}^{-1}=W$. Since $g\in G$, $U$ and $W$ are conjugate in $G$.

Conversely, assume that $U$ is conjugate to $W$ in $G$. Then there is some $g\in G$ such that $\mathit{𝑔𝑈}{g}^{-1}=W$.

Note first that $P$ is a subgroup of $N_G(U)$, since $P=N_P(U)=P\cap N_G(U)$.

Moreover, $g^{-1}\mathit{𝑃𝑔}\le N_G(U)$: if $q\in g^{-1}\mathit{𝑃𝑔}$, then $q=g^{-1}\mathit{𝑝𝑔}$ for some $p\in P$, and

so $q\in N_G(U)$. This shows that $g^{-1}\mathit{𝑃𝑔}\le N_G(U)$, where $|g^{-1}\mathit{𝑃𝑔}|=|P|$.

So $P$ and $g^{-1}\mathit{𝑃𝑔}$ are Sylow $p$-subgroups of $N_G(U)$. By the second part of Sylow’s Theorem, these two subgroups are conjugate in $N_G(U)$, so there is some $h\in N_G(U)$ such that

Put $k=\mathit{𝑔h}$. Then $\mathit{𝑘𝑃}{k}^{-1}=P$, so $k\in N_G(P)$.

Moreover,

This proves that $U$ and $W$ are conjugate in $N_G(P)$.

If $U$ and $W$ are normal subsets of a Sylow $p$-subgroup $P$ of $G$ then $U$ is conjugate to $W$ in $G$ if and only if $U$ is conjugate to $W$ in $N_G(P)$.

Let $u,w$ be elements of $Z=Z(P)$. Then $U=\{u\}$ and $W=\{w\}$ are normal subsets of the Sylow $p$-subgroup $P$, and $u,w$ are conjugate in $G$ if and only if $U=\{u\}$ and $W=\{w\}$ are conjugate sets in $G$.

By the first part, this is true if and only if $U$ and $W$ are conjugate in $N_G(P)$, if and only if $u$ and $w$ are conjugate in $N_G(P)$.

Two elements in the center of $P$ are conjugate in $G$ if and only if they are conjugate in $N_G(P)$. □