Frattini's argument

If $G$ is a finite group with normal subgroup $H$, and if $P$ is a Sylow p-subgroup of $H$, then

: $G\; =\; N\_G(P)H,$

where $N\_G(P)$ denotes the normalizer of $P$ in $G$, and $N\_G(P)H$ means the product of group subsets.

The group $P$ is a Sylow $p$-subgroup of $H$, so every Sylow $p$-subgroup of $H$ is an $H$-conjugate of $P$, that is, it is of the form $h^\{-1\}Ph$ for some $h\; \backslash in\; H$ (see Sylow theorems). Let $g$ be any element of $G$. Since $H$ is normal in $G$, the subgroup $g^\{-1\}Pg$ is contained in $H$. This means that $g^\{-1\}Pg$ is a Sylow $p$-subgroup of $H$. Then, by the above, it must be $H$-conjugate to $P$: that is, for some $h\; \backslash in\; H$

: $g^\{-1\}Pg\; =\; h^\{-1\}Ph,$

and so

: $hg^\{-1\}Pgh^\{-1\}\; =\; P.$

Thus

: $gh^\{-1\}\; \backslash in\; N\_G(P),$

and therefore $g\; \backslash in\; N\_G(P)H$. But $g\; \backslash in\; G$ was arbitrary, and so $G\; =\; HN\_G(P)\; =\; N\_G(P)H.\backslash \; \backslash square$

• Frattini's argument can be used as part of a proof that any finite nilpotent group is a direct product of its Sylow subgroups.
• By applying Frattini's argument to $N\_G(N\_G(P))$, it can be shown that $N\_G(N\_G(P))\; =\; N\_G(P)$ whenever $G$ is a finite group and $P$ is a Sylow $p$-subgroup of $G$.
• More generally, if a subgroup $M\; \backslash leq\; G$ contains $N\_G(P)$ for some Sylow $p$-subgroup $P$ of $G$, then $M$ is self-normalizing, i.e. $M\; =\; N\_G(M)$.

• [https://proofwiki.org/wiki/Frattini%27s_Argument Frattini's Argument on ProofWiki]

{{Reflist}}

• {{Cite book

| last = Hall

| first = Marshall

| author-link = Marshall Hall (mathematician)

| title = The theory of groups

| publisher = Macmillan

| date = 1959

| location = New York, N.Y.

}} (See Chapter 10, especially Section 10.4.)

{{DEFAULTSORT:Frattini's Argument}}

Category:Lemmas in group theory

Category:Articles containing proofs