Sylow theorems

If $|G|=p^{a}m$ with $p\nmid m$, then $G$ has subgroups of order $p^a$, every $p$-subgroup lies in one, and all such Sylow subgroups are conjugate. Their number satisfies $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid m$.

If $H$ is a normal subgroup of a finite group $G$ and $P$ a Sylow subgroup of $H$, thenNormality puts every $g^{-1}Pg$ inside $H$, and Sylow conjugacy there gives $g^{-1}Pg=h^{-1}Ph$. Hence $g{h}^{-1}$ normalizes $P$, giving the factorization. This reduces questions about $G$ to a normaliser and its normal subgroup.

For distinct primes $p,q$, the Sylow theorems force a proper nontrivial normal subgroup in any group of order $p^{2}q$. If $p>q$, the Sylow $p$-subgroup is unique. If $p<q$ and both Sylow counts were nontrivial, the count congruences force $(p,q)=(2,3)$. In order twelve, four Sylow $3$-subgroups exhaust eight nonidentity elements; only three remain for a Sylow $2$-subgroup, making that subgroup unique. Thus the group is not a simple group.

A Sylow $p$-subgroup of a finite group has the largest power of $p$ dividing the group order.

Let $p>q$ be prime numbers. A nonabelian group of order $pq$ has a unique Sylow $p$-subgroup and exactly $p$ Sylow $q$-subgroups. Hence the Sylow count satisfies $p\equiv 1(\mathrm{mod}q)$, so $q\mid p-1$.

Up to isomorphism, the unique nonabelian group of order $21$ isIts nonidentity elements of $C_7$ form two conjugacy classes of size three, and the other two $C_7$ cosets form conjugacy classes of size seven.

A group, or one of its Sylow subgroups, acts on the set of Sylow subgroups by conjugation.

A finite nonabelian simple group embeds in $A_{n_p}$, where $n_p$ is its number of Sylow subgroups for a prime dividing its order. It is not a p-group by the nontrivial-centre property. Thus its Sylow subgroups are proper and cannot be normal, so $n_p>1$. The conjugation action is nontrivial, and its normal kernel of a group homomorphism must therefore be trivial. The sign of this faithful permutation action must be trivial, since a nontrivial sign homomorphism would inject the group into a group of order two. Lagrange's theorem consequently gives $|G|\mid n_p!/2$.

The Sylow $2$-subgroups of $S_3$ are its three transposition subgroups, while its unique Sylow $3$-subgroup is $A_3$. The three Sylow $2$-subgroups of $S_4$ are dihedral groups of order eight and are the normalizers of the three cyclic subgroups generated by inverse pairs of $4$-cycles. The five Sylow $2$-subgroups of $A_5$ are Klein four-groups, one fixing each letter.

If $P$ is a Sylow $p$-subgroup of $G$ and $Q$ is any $p$-subgroup, let $Q$ act on $G/P$. Since $p$ does not divide $|G/P|$, some coset $gP$ is fixed. Hence $g^{-1}Qg\le P$, or equivalently $Q\le gP{g}^{-1}$.

Suppose a group acts faithfully and transitively on seven points, every point stabilizer is $S_4$, and every two-point stabilizer is a Klein four-group. Then the group has order $168$ andCount pairs consisting of a point and a Sylow subgroup fixing it for $p=2,3$; the two-point stabilizer ensures uniqueness. For $p=7$, a normal Sylow subgroup would force the group into its order-$42$ normalizer in $S_7$.

Suppose a finite group $G$ has no index-two subgroup, $P$ is a Sylow $2$-subgroup, $H$ has index two in $P$, and $x\in G$ has order two. The sign of the action on $G/H$ defines a homomorphism $G\to \{\pm 1\}$ and is therefore trivial. Since $[G:H]\equiv 2(\mathrm{mod}4)$, a fixed-point-free involution would be an odd number of transpositions. Thus $x$ fixes a coset $gH$, equivalently $g^{-1}xg\in H$.