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Perhaps this is not a direct answer to your question, but one of my favorite examples of $p$-groups entering into the theory of general finite groups (except of course for Sylow theory itself!) occurs in a theorem due to Frobenius. This theorem states:

Let $P$ be a Sylow $p$-subgroup of a finite group $G$. Then the following are equivalent:

(i) $G$ has a normal $p$-complement.

 

(ii) $N_G(U)$ has a normal $p$-complement for all $p$-subgroups $U\subseteq G$ with $U>1$.

 

(iii) $N_G(U)/C_G(U)$ is a $p$-group for all $p$-subgroups $U\subseteq G$.

 

(iv) There is no fusion in $P$.

(Here $C_G(U)$ and $N_G(U)$ denote the centralizer and normalizer of the subgroup $U\subseteq G$, respectively. The quotient $N_G(U)/C_G(U)$ makes sense, of course, since $C_G(U)$ is normal in $N_G(U)$. Also, to say that there is no fusion in $P$ is to say that any pair of elements in $P$ that are $G$-conjugate are already $P$-conjugate.)

A remarkable theorem of Thompson asserts that, if $p\neq 2$ in Frobenius' theorem, to verify that $G$ has a normal $p$-complement is tantamount to checking that $N_G(U)$ has a normal $p$-complement for just two particular $p$-subgroups $U\subseteq G$. The two subgroups $U\subseteq G$ whose normalizers need to be checked are $Z(P)$ and $J(P)$ where $P$ is a fixed Sylow $p$-subgroup of $G$ and $Z(P)$ and $J(P)$ are the center and Thompson subgroup of $P$, respectively.

Perhaps this is not a direct answer to your question, but one of my favorite examples of $p$-groups entering into the theory of general finite groups (except of course for Sylow theory itself!) occurs in a theorem due to Frobenius. This theorem states:

Let $P$ be a Sylow $p$-subgroup of a finite group $G$. Then the following are equivalent:

(i) $G$ has a normal $p$-complement.

 

(ii) $N_G(U)$ has a normal $p$-complement for all $p$-subgroups $U\subseteq G$ with $U>1$.

 

(iii) $N_G(U)/C_G(U)$ is a $p$-group for all $p$-subgroups $U\subseteq G$.

 

(iv) There is no fusion in $P$.

(Here $C_G(U)$ and $N_G(U)$ denote the centralizer and normalizer of the subgroup $U\subseteq G$, respectively. The quotient $N_G(U)/C_G(U)$ makes sense, of course, since $C_G(U)$ is normal in $N_G(U)$. Also, to say that there is no fusion in $P$ is to say that any pair of elements in $P$ that are $G$-conjugate are already $P$-conjugate.)

A remarkable theorem of Thompson asserts that, if $p\neq 2$ in Frobenius' theorem, to verify that $G$ has a normal $p$-complement is tantamount to checking that $N_G(U)$ has a normal $p$-complement for just two particular $p$-subgroups $U\subseteq G$. The two subgroups $U\subseteq G$ whose normalizers need to be checked are $Z(P)$ and $J(P)$ where $P$ is a fixed Sylow $p$-subgroup of $G$ and $Z(P)$ and $J(P)$ are the center and Thompson subgroup of $P$, respectively.

Perhaps this is not a direct answer to your question, but one of my favorite examples of $p$-groups entering into the theory of general finite groups (except of course for Sylow theory itself!) occurs in a theorem due to Frobenius. This theorem states:

Let $P$ be a Sylow $p$-subgroup of a finite group $G$. Then the following are equivalent:

(i) $G$ has a normal $p$-complement.

(ii) $N_G(U)$ has a normal $p$-complement for all $p$-subgroups $U\subseteq G$ with $U>1$.

(iii) $N_G(U)/C_G(U)$ is a $p$-group for all $p$-subgroups $U\subseteq G$.

(iv) There is no fusion in $P$.

(Here $C_G(U)$ and $N_G(U)$ denote the centralizer and normalizer of the subgroup $U\subseteq G$, respectively. The quotient $N_G(U)/C_G(U)$ makes sense, of course, since $C_G(U)$ is normal in $N_G(U)$. Also, to say that there is no fusion in $P$ is to say that any pair of elements in $P$ that are $G$-conjugate are already $P$-conjugate.)

A remarkable theorem of Thompson asserts that, if $p\neq 2$ in Frobenius' theorem, to verify that $G$ has a normal $p$-complement is tantamount to checking that $N_G(U)$ has a normal $p$-complement for just two particular $p$-subgroups $U\subseteq G$. The two subgroups $U\subseteq G$ whose normalizers need to be checked are $Z(P)$ and $J(P)$ where $P$ is a fixed Sylow $p$-subgroup of $G$ and $Z(P)$ and $J(P)$ are the center and Thompson subgroup of $P$, respectively.

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Amitesh Datta
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Perhaps this is not a direct answer to your question, but one of my favorite examples of $p$-groups entering into the theory of general finite groups (except of course for Sylow theory itself!) occurs in a theorem due to Frobenius. This theorem states:

Let $P$ be a Sylow $p$-subgroup of a finite group $G$. Then the following are equivalent:

(i) $G$ has a normal $p$-complement.

(ii) $N_G(U)$ has a normal $p$-complement for all $p$-subgroups $U\subseteq G$ with $U>1$.

(iii) $N_G(U)/C_G(U)$ is a $p$-group for all $p$-subgroups $U\subseteq G$.

(iv) There is no fusion in $P$.

(Here $C_G(U)$ and $N_G(U)$ denote the centralizer and normalizer of the subgroup $U\subseteq G$, respectively. The quotient $N_G(U)/C_G(U)$ makes sense, of course, since $C_G(U)$ is normal in $N_G(U)$. Also, to say that there is no fusion in $P$ is to say that any pair of elements in $P$ that are $G$-conjugate are already $P$-conjugate.)

A remarkable theorem of Thompson asserts that, if $p\neq 2$ in Frobenius' theorem, to verify that $G$ has a normal $p$-complement is tantamount to checking that $N_G(U)$ has a normal $p$-complement for just two particular $p$-subgroups $U\subseteq G$. The two subgroups $U\subseteq G$ whose normalizers need to be checked are $Z(P)$ and $J(P)$ where $P$ is a fixed Sylow $p$-subgroup of $G$ and $Z(P)$ and $J(P)$ are the center and Thompson subgroup of $P$, respectively.