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For order $48=2^4\cdot 3$, the Sylow theorems give $n_2\mid 3$ and $n_2\equiv 1(\mathrm{mod}2)$, hence $n_2=1$ or $3$. If $n_2=1$, the Sylow subgroup of order $16$ is a nontrivial proper normal subgroup. If $G$ were simple in the other case, it would be nonabelian (a finite abelian simple group has prime order), and the preceding result would force $48\mid 3!/2=3$, impossible. Thus every group of order 48 is not simple.

An example with no normal Sylow subgroup for $p=2$ isIn $S_4$, $P=⟨(1234),(13)⟩$ is a dihedral group of order $8$: conjugation by $(13)$ inverts $(1234)$. Thus $P\times C_2$ has order $16$ and is a Sylow subgroup of $G$. The only transpositions in $P$ are $(13)$ and $(24)$; conjugating $(13)$ by $(23)$ gives $(12)\notin P$. Therefore $P$ and $P\times C_2$ are not normal. By conjugacy, no other Sylow subgroup for $p=2$ is normal either.

Write $|G|=p^{n}m$ with $p\nmid m$. The Sylow theorems state:

Let $|G|=pq$ with primes $p>q$ and $G$ nonabelian. The Sylow theorems give $n_p\mid q$ and $n_p\equiv 1(\mathrm{mod}p)$, so $n_p=1$ and the Sylow $p$-subgroup $P$ is normal. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal and $G$ is their abelian direct product, a contradiction. Hence $n_q=p$, and

Because $G/P$ is abelian, the commutator subgroup $G^{\prime }$ lies in $P$. It is nontrivial because $G$ is nonabelian, and $|P|=p$ is prime, so $G^{\prime }=P$. Thereforeand $G$ has exactly $q$ linear characters.

Every nonlinear irreducible degree divides $pq$. Its square is at most $pq$, so the only possibility is degree $q$. If there are $b$ such characters, the sum of squares of irreducible degrees givesThe number of irreducible characters equals the number of conjugacy classes, soThese are the irreducible characters of a nonabelian group of order p q.

The Sylow theorems state that if $|G|=p^{a}m$ with $p\nmid m$, then:

Interchange the two primes if necessary so that $p<q$. By the Sylow theorems, the number $n_q$ of Sylow $q$-subgroups satisfiesIf $n_q\ne 1$, then $n_q$ is $p$ or $p^2$. The first is impossible because $1<p<q$. The second would implySince $q>p$ is prime, it cannot divide $p-1$, so it must divide $p+1$. But $q>p$ then forces $q=p+1$, impossible because $p$ and $q$ are both odd. Therefore $n_q=1$. The unique Sylow $q$-subgroup is a nontrivial proper normal subgroup, and hence

Take $g\in G$. Since $K$ is a normal subgroup, $gP{g}^{-1}\le K$; it has the same order as $P$, so it is also a Sylow subgroup of $K$. The conjugacy assertion in the Sylow theorems supplies $k\in K$ such thatConsequently $kg\in N_G(P)$, and thereforeThus $G\subseteq N_G(P)K$, while the reverse inclusion is immediate because both factors are subgroups of $G$. Hence

For subgroups $H,P\le G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the double cosets $HxP$ partition $G$.

Let $H$ act by left multiplication on the set of left cosets $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer isThe orbit-stabilizer theorem therefore gives

Suppose $P$ is a Sylow subgroup of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xP{x}^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xP{x}^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the Sylow subgroup of a subgroup from double cosets argument.

To count the general linear group over a finite field $G{L}_n({\mathbb{F}}_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. ConsequentlyNone of the factors $p^j-1$ is divisible by $p$, so the upper unitriangular group is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The permutation matrices form a subgroup isomorphic to the symmetric group $S_n$.

By Cayley theorem, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $G{L}_{|G|}({\mathbb{F}}_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.

The counting part of the Sylow theorems says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies

Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts giveso $n_p=1$ and the Sylow $p$-subgroup is a normal subgroup. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their direct product and is abelian. In the nonabelian case one must therefore have $n_q=p$, whenceor equivalently $q\mid p-1$, as recorded by nonabelian group of order pq.

A Sylow subgroup of a finite group $G$ is a subgroup whose order is the largest power of a prime $p$ dividing $|G|$. The Sylow theorems say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv 1(\mathrm{mod}p)$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every orbit other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\le N_G(Q)$. They are Sylow subgroups of this normalizer, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv 1(\mathrm{mod}p)$.

Now let $H<A_n$ have index $m$ with $1<m<n$. The coset action gives a homomorphismIts kernel is a normal subgroup of the simple group $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfiesand simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the sign homomorphism $S_6\to \{\pm 1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus

Let $Q$ act by left multiplication on the set $G/P$ of left cosets. Every orbit has size a power of $p$, whileis not divisible by $p$. Therefore at least one orbit has size one. If $gP$ is fixed, then $qgP=gP$ for every $q\in Q$, so $g^{-1}Qg\le P$, equivalentlyThis is Sylow containment from a coset fixed point.

The remaining Sylow theorems state that Sylow $p$-subgroups exist, that any two are conjugate, and that their number $n_p$ satisfies

Let $|G|=p^{a}m$ with $p\nmid m$. The Sylow theorems state: