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 $n_p\mid m$.

Let $p<q$ be distinct prime numbers. Every group of group order $p^2{q}^2$ is Abelian if and only if $p$ does not divide $q^2-1$.

For sufficiency, the Sylow theorems make the number of Sylow p-subgroups one of $1,q,q^2$, congruent to one modulo $p$. The divisibility hypothesis excludes both nontrivial choices, so that subgroup is normal. The number of Sylow subgroups for $q$ is one of $1,p,p^2$, congruent to one modulo $q$. The choice $p$ is impossible because $p<q$. If $q$ divided $p^2-1=(p-1)(p+1)$, primality and $q>p$ would force $q=p+1$, hence $(p,q)=(2,3)$; this pair violates the hypothesis. Thus the other Sylow subgroup is normal too. Their intersection is trivial, and their group commutators lie in that intersection, so the group is their direct product of groups. A group of prime-square order is abelian: its nontrivial center of a group is either the whole group or has cyclic quotient, and a cyclic central quotient also forces commutativity. Both factors are therefore abelian.

For necessity, suppose $p\mid q^2-1$. The multiplicative group of a finite field is cyclic, so ${\mathbb{F}}_{q^2}^{\times }$ contains $\alpha$ of order $p$. Let the cyclic group $C_{p^2}$ act on the additive group of ${\mathbb{F}}_{q^2}$ by mapping its generator to multiplication by $\alpha$. This is a well-defined nontrivial group action, since ${\alpha }^{p^2}=1$. The resulting semidirect product has order $p^2{q}^2$ and is nonabelian, because the generator does not commute with a nonzero field element. Thus a nonabelian group of this order exists exactly when $p\mid q^2-1$.

A Sylow basis consists of one Sylow subgroup for each prime dividing the group order, with any two of them permutable as subgroups. Permutable means $P_p{P}_q=P_q{P}_p$, not elementwise commutation. Their product over any subset of primes is a Hall subgroup. Every finite soluble group has such a basis, and all bases are simultaneously conjugate. Induction over an elementary abelian minimal normal subgroup proves these facts using coprime splitting over an elementary abelian normal subgroup: lift the other-prime quotient basis into a complement, and lift the characteristic-prime Sylow subgroup by taking its full preimage.

Let $V$ be an elementary abelian $r$-group and let $r\nmid |H|$, with $H$ soluble. In each Sylow basis of $V⋊H$, $V$ is the unique Sylow $r$-subgroup and the product of the remaining Sylow subgroups is a complement to $V$. All complements are $V$-conjugate, and the stabilizer of $H$ under this conjugation is its fixed-vector space $C_V(H)$. Each complement has $b(H)$ Sylow bases, proving the formula. For the permutation module $V={\mathbb{F}}_5^4$ of $S_4$, the fixed-vector space is the one-dimensional constant-coordinate subspace, so the number is $5^3\cdot 12=1500$.

Let $|G|=p^{n}r$ with $p\nmid r$. The coefficient of $X^{p^n}$ in $(1+X{)}^{p^{n}r}$ is $r$ modulo $p$, because $(1+X{)}^{p^{n}r}=(1+X^{p^n}{)}^r$ over the field of $p$ elements. Thus the number of subsets of size $p^n$ is not divisible by $p$. Left translation of $G$ on those subsets has an orbit of index prime to $p$. Its stabilizer acts freely on its subset, so its order divides $p^n$, while its prime-to-$p$ index forces its order to be $p^n$. This proves existence of a Sylow subgroup without an induction using Cauchy's theorem.

The Sylow theorems give either one or eight Sylow subgroups of order seven. In the first case that subgroup is normal. In the second, their six nonidentity elements are pairwise disjoint, accounting for 48 elements. Only eight elements remain, including the identity, and every Sylow subgroup of order eight lies entirely among them. Thus there is exactly one, and it is a normal subgroup.

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.

In the alternating group $A_6$, of order $360$, a Sylow subgroup of order five is generated by a five-cycle. There are $6\cdot 4!=144$ five-cycles, four in each such subgroup, giving $36$ subgroups. A Sylow subgroup of order nine is generated by disjoint three-cycles, hence is $C_3\times C_3$. The unordered partitions of six letters into two triples give all such subgroups by Sylow conjugacy, and their single three-cycles recover the partition uniquely. There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$ partitions.

Suppose distinct Sylow subgroups for $p$ in a finite group have trivial intersections. Conjugation by one of them, $P$, fixes $P$. The stabilizer of any other $Q$ is $H=P\cap N_G(Q)$. Since $H$ normalizes $Q$, the product $HQ$ is a subgroup of $p$-power order; maximality of $Q$ gives $H\le Q$. Thus $H\le P\cap Q=1$. All other orbits have size $|P|$, proving the congruence.

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.

For a finite group and a Sylow subgroup $P$, conjugation by $P$ on all Sylow subgroups has stabilizer $P\cap Q$ at $Q$: the normal Sylow subgroup $Q$ of $N_G(Q)$ contains every $p$-subgroup of that normalizer. Thus the orbit length is $[P:P\cap Q]$. If every $Q\ne P$ has this index at least $p^a$, each such orbit length is divisible by $p^a$, whereas $P$ is a singleton orbit. Summing orbit lengths proves the displayed stronger integer congruence.

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 $g^{-1}xg\in H$.