// Workers AI · dad joke modeWhat did Frattini's argument say at the party? It fractured the conversation.

In group theory, a branch of mathematics, Frattini's argument is an important lemma in the structure theory of finite groups. It is named after Giovanni Frattini, who used it in a paper from 1885 when defining the Frattini subgroup of a group. The argument was taken by Frattini, as he himself admits, from a paper of Alfredo Capelli dated 1884.

Frattini's argument

Statement

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.

Proof

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 $h\in H$ (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 $h\in H$

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

and so

$hg^{-1}Pgh^{-1}=P.$

Thus

$gh^{-1}\in N_{G}(P),$

and therefore $g\in N_{G}(P)H$. But $g\in G$ was arbitrary, and so $G=HN_{G}(P)=N_{G}(P)H.\ \square$

Applications

• 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 $M\leq G$ contains $N_{G}(P)$ for some Sylow $p$-subgroup $P$ of $G$, then $M$ is self-normalizing, i.e. $M=N_{G}(M)$.

External links

• Frattini's Argument on ProofWiki

Further reading

• Hall, Marshall (1959). The theory of groups. New York, N.Y.: Macmillan. (See Chapter 10, especially Section 10.4.)