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The upper unitriangular group is generated by adjacent elementary transvection matrices $I+s{E}_{i,i+1}$. Commuting a matrix $X$ with each such generator forces every entry above the diagonal to vanish except possibly its $(1,n)$ entry: comparing $X{E}_{i,i+1}$ with $E_{i,i+1}X$ kills the entries above position $(i,i)$ in column $i$ and to the right of position $(i+1,i+1)$ in row $i+1$. The surviving matrices $I+t{E}_{1n}$ do commute with the entire group. In particular the group center has order $q$; it distinguishes $U_4(2)$ from $U_3(4)$ despite their common order $64$.

If every nonidentity element of a finite group has order equal to a prime $p$, its order is a power of $p$. For any prime divisor $q$ of the group order, Cauchy theorem for groups gives an element of order $q$, forcing $q=p$. This statement does not imply that the group is abelian; odd-characteristic upper unitriangular groups provide counterexamples.

Write $x_{ij}(t)=I+t{E}_{ij}$ for $i\ne j$ and $t\in {\mathbb{F}}_q$. When $t\ne 0$ these are transvections, and their inverses are $x_{ij}(-t)$. Row addition by such an elementary transvection matrix preserves the determinant. It also realizes signed interchanges and compensating diagonal scalings: in any two coordinate positions,Thus elimination can reduce any invertible matrix to diagonal form using only row additions and signed interchanges, all products of transvections. For a diagonal matrix with determinant one, $\mathrm{diag}(d_1,\dots ,d_n)$, use the displayed two-coordinate diagonal matrices in positions $(i,n)$ with parameter $d_i$, for $1\le i<n$. Their product has diagonal entries $d_1,\dots ,d_{n-1},{\prod }_{i<n}{d}_i^{-1}=d_n$. Hence the residual diagonal matrix is also a product of transvections, and reversing elimination provesThis proves transvections generate the special linear group, including the small fields.

For perfectness, use the convention $[x,y]=xy{x}^{-1}{y}^{-1}$ in the following calculations. If $n\ge 3$ and $i,j,k$ are distinct, direct matrix multiplication givesTaking $b=1$ puts every elementary generator in the commutator subgroup. Thus $S{L}_n(q)$ is perfect. When $n=2$ and $q>3$, choose $s\in {\mathbb{F}}_q^{\times }$ with $s^2\ne 1$; such an element exists because a degree-two polynomial has at most two roots and $q-1>2$. For $D=\mathrm{diag}(s,s^{-1})$,Both coefficients are nonzero, so varying $t$ supplies every upper and lower elementary generator as a commutator. ThereforeThe exceptions really fail: $S{L}_2(2)\cong S_3$ has a nontrivial sign quotient, and $S{L}_2(3)$ has quotient $PS{L}_2(3)\cong A_4$, which has the nontrivial Abelian quotient $A_4/V_4\cong C_3$. For the latter identification, the faithful projective action embeds the order-twelve group $PS{L}_2(3)$ as an index-two subgroup of $S_4$, necessarily $A_4$.

The Iwasawa simplicity lemma states that, in a faithful primitive group action, if a point stabilizer has an Abelian normal subgroup whose conjugates generate the whole group, then every nontrivial normal subgroup contains the commutator subgroup. In particular a nontrivial perfect group with this property is simple. The lemma is applied here, as allowed, without assuming the desired simplicity theorem.

Let $S{L}_n(q)$ act on the one-dimensional subspaces of ${\mathbb{F}}_q^n$. This action is two-transitive: a linear map taking two specified distinct lines to two others can be made determinant one by rescaling one of the target basis vectors, without changing either target line. Its kernel consists exactly of scalar matrices. Indeed fixing all lines makes a matrix diagonal in any fixed basis, and fixing the lines through $e_i+e_j$ makes all its diagonal entries equal. The scalar matrices in $S{L}_n(q)$ are $aI$ with $a^n=1$. They also form its group center, since a central matrix commutes with every elementary generator, forcing all off-diagonal entries to vanish and all diagonal entries to agree. Their number is $\gcd (n,q-1)$. Consequently the quotient $PS{L}_n(q)$ acts faithfully and primitively.

Fix a line $L=⟨v⟩$ and considerHere $vf$ is the rank-at-most-one map $z\mapsto f(z)v$. These matrices have determinant one, and $(vf)(vh)=0$, so $T_L$ is Abelian under addition of the functionals. If $gL=L$, then $gv=av$ and $g(I+vf)g^{-1}=I+av(f{g}^{-1})\in T_L$. Thus $T_L$ is normal in the line stabilizer. Its image in $PS{L}_n(q)$ is a nontrivial Abelian normal subgroup of the point stabilizer. Conjugates of these groups include every elementary transvection, hence generate $PS{L}_n(q)$. Perfectness passes to a quotient. The Iwasawa simplicity lemma therefore givesIn those exceptions the projective groups are $S_3$ and $A_4$, as above.

For $q=4$, counting columns gives $|S{L}_2(4)|=4(4^2-1)=60$. Its scalar group center is trivial, so $PS{L}_2(4)$ has order sixty. Its faithful action on the five points of the projective line embeds it in $S_5$. Because it is perfect, its sign homomorphism is trivial and its image lies in $A_5$. Equal orders then giveFor $q=5$, use an explicit subgroup to obtain a five-point action. In $S{L}_2(5)$ setAll three have determinant one. Matrix multiplication gives $i^2=j^2=-I$, $ij=-ji$, and $r^3=I$. The eight distinct matrices $\{\pm I,\pm i,\pm j,\pm ij\}$ form a quaternion group. Moreover,Thus $r$ normalizes that quaternion group and has order three. Their generated subgroup has order $24$ and contains the scalar group center $\{\pm I\}$. Its image in $PS{L}_2(5)$ has order twelve, hence index five since $|PS{L}_2(5)|=5(25-1)/2=60$. The resulting coset action is nontrivial and, by the simplicity already proved, faithful. Perfectness again puts its image in $A_5$, and equal orders implyThis is the quaternion construction of an index-five subgroup of PSL2 over F5; it avoids assuming a classification of groups of order sixty.

For the last comparison, the order of a general linear group over a finite field comes from choosing independent columns:The determinant map onto ${\mathbb{F}}_q^{\times }$ has kernel $S{L}_n(q)$, and the scalar quotient divides by $\gcd (n,q-1)$. HenceThe highest power of two dividing either order is $64$. Their Sylow $2$-subgroups are respectively the upper unitriangular groups $U_4(2)$ of order $2^6$ and the image of $U_3(4)$ of order $4^3$. The latter image is isomorphic to $U_3(4)$ because its intersection with the odd-order scalar group center is trivial.

The center of an upper unitriangular group can be computed directly here. Commuting $X\in U_4(2)$ with $I+E_{12}$ forces $X_{23}=X_{24}=0$; commuting with $I+E_{23}$ forces $X_{12}=X_{34}=0$; commuting with $I+E_{34}$ forces $X_{13}=0$. Only $X_{14}$ can remain, and those matrices commute with the whole group. Thus $Z(U_4(2))=\{I+t{E}_{14}:t\in {\mathbb{F}}_2\}$ has order two. In $U_3(4)$, commutation with $I+E_{12}$ and $I+E_{23}$ similarly forces $X_{23}=X_{12}=0$, leaving the arbitrary entry $X_{13}\in {\mathbb{F}}_4$. Its group center has order four. An isomorphism of the whole groups would send a Sylow subgroup to a Sylow subgroup and preserve its group center order, which is impossible. Therefore

Write a matrix in the upper unitriangular group as $M=I+A$. Here $A^3=0$, and its only potentially nonzero entry in $A^2$ is $(A^2{)}_{13}=ab$. The binomial theorem therefore gives the unitriangular matrix power formulaFor an odd prime number $p$, both $p$ and $\left(\genfrac{}{}{0ex}{}{p}{2}\right)=p(p-1)/2$ vanish in the finite field ${\mathbb{F}}_p$. Thus $M^p=I$. A nonidentity $M$ has order dividing the prime $p$ (by the same exponent-division argument as in part (i)), and so has order exactly $p$.

For $p=2$, take $a=b=1$ and $x=0$. Then $M^2=I+A^2\ne I$, while $M^4=(I+A^2{)}^2=I$. This element has order $4$, not $2$. Hence every nonidentity element has order $p$ exactly for odd $p$. The counterexample at $p=2$ accounts for the quadratic term in the matrix power formula.

An example is the unitriangular group of degree three over F3, consisting ofThis is an upper unitriangular group of order $3^3=27$. Direct matrix multiplication givesThe formula gives closure, identity $M(0,0,0)$, and inverse $M(-a,-b,ab-c)$; associativity follows from matrix multiplication. The two elements $M(1,0,0)$ and $M(0,1,0)$ do not commute: their products have respectively $c=1$ and $c=0$.

Every element has the form $I+N$ with $N^3=0$. Since the finite field ${\mathbb{F}}_3$ has characteristic three, the binomial expansion givesHence every nonidentity element has order exactly three. This is a nonabelian finite group with exponent of a finite group equal to three.

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.

Let $N$ be the subgroup whose upper-left block is $I_2$. It is a finitely generated nilpotent subgroup of the integer upper unitriangular group $U{T}_4(\mathbb{Z})$ and is normal in $L$. The block-diagonal matrixgenerates an infinite cyclic quotient, soFinitely generated nilpotent groups are polycyclic, and an extension of polycyclic groups is polycyclic. Hence $L$ is polycyclic.

Inside $L$, retain only $t$ and the entries in positions $(1,3)$ and $(2,3)$. They form a subgroupThe characteristic polynomial of $A$ is ${\lambda }^2-5\lambda +1$, so its eigenvalues areOne has modulus greater than one, and the resulting semidirect product has exponential growth. Every finitely generated virtually nilpotent group has polynomial growth, as does each of its finitely generated subgroups. Therefore $L$ cannot be virtually nilpotent.

The upper unitriangular group $U{T}_3({\mathbb{F}}_3)$ consists of $M(a,b,c)=\left(\begin{array}{ccc}1& a& c\\ 0& 1& b\\ 0& 0& 1\end{array}\right)$ for $a,b,c\in {\mathbb{F}}_3$. Its multiplication is $M(a,b,c)M(a^{\prime },b^{\prime },c^{\prime })=M(a+a^{\prime },b+b^{\prime },c+c^{\prime }+a{b}^{\prime })$. The elements $M(1,0,0)$ and $M(0,1,0)$ do not commute. Yet every element is $I+N$ with $N^3=0$, so characteristic three gives $(I+N{)}^3=I$. It is a nonabelian finite group of order $27$ whose nonidentity elements all have order three.