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For an invertible fundamental matrix, differentiating $Y_{\lambda }=AY$ in $z$ and $Y_z=BY$ in $\lambda$ gives $Y_{\lambda z}=(A_z+AB)Y$ and $Y_{z\lambda }=(B_{\lambda }+BA)Y$. Thus the necessary and sufficient compatibility identity isHere both derivatives must use the actual deformation variable $z$; the printed $t$ has not been defined. The isomonodromic deformation identity can be checked as a polynomial in $\lambda$, without assuming the conclusion.

First retain the coefficients actually printed in the PDF. Write $p=u^{\prime }$, $d=w+z/2$, and introduce the two residualsDirect matrix multiplication and differentiation giveFor example, the quadratic lower entry is the difference between $[D,B_0{]}_{21}=2w/u$ and the contribution $(A_1{)}_{21}=-zw/u$; it does not cancel. The constant lower entry follows by differentiating $2u^{\prime }w/u^2$ and collecting it as $2wH/u^2+2u^{\prime }K/u^2$. These entries also provide independent checks of the two most important cancellations.

Work on an open connected deformation domain where $u\ne 0$, as the matrices require. The quadratic entry forces $(2-z)w=0$ throughout the domain, hence $w\equiv 0$ by continuity, including a possible point $z=2$. All remaining entries then vanish exactly when $u^{\prime \prime }+(z/2)u=0$. Thus the literal answer isThis is a rescaled Airy function equation: with $\xi =-2^{-1/3}z$, the equation is $u_{\xi \xi }-\xi u=0$, so $u=c_1\mathrm{Ai}(\xi )+c_2\mathrm{Bi}(\xi )$ on a domain avoiding its zeros. In particular, choose the local solution with $u(1)=u^{\prime }(1)=1$ and $w=0$. It gives compatible matrices but violates the printed $w=u^{\prime }$. This proves that the requested assertion is false for the original coefficients, independently of the undefined $t/z$ in the printed differential equation.

For completeness, use the explicitly corrected lower coefficient $(A_1{)}_{21}=-2w/u$ from the preceding solution. The same multiplication gives the much simpler residualConsequently compatibility is equivalent to $K=H=0$. The first equation integrates without dividing by $w$:The constant of integration $C$ therefore satisfies $w=C{u}^2$, and the second equation becomesConversely these two equations give $K=H=0$, and substitution makes every residual entry vanish, proving sufficiency as well as necessity. This is the quadratic spectral Lax pair for zero-parameter Painlevé II. For $C\ne 0$, choose constants $b,a$ with $b^3=-2$ and $C{a}^2{b}^2=-2$, set $z=bx$, $u=aq(x)$, and obtain $q^{\prime \prime }=2q^3+xq$, the zero-parameter Painlevé II equation. For $C=0$ it reduces to the Airy function equation. Even with this correction, the conclusions are $w=C{u}^2$ and the coefficient $z/2$, rather than the incompatible expressions printed in the question.

Suppose $K=\ker \phi$ for a group homomorphism $\phi :G\to H$. Its identity belongs to $K$, and for $k,l\in K$ we have $\phi (k{l}^{-1})=1$, so it is a subgroup. Moreover, for $g\in G$ and $k\in K$,Hence $gK{g}^{-1}\subseteq K$; applying this also to $g^{-1}$ gives equality, so $K$ is a normal subgroup.

Conversely, if $K$ is a normal subgroup, multiplication of cosets in the quotient group $G/K$ is well-defined. The map $g\mapsto gK$ is a group homomorphism and has kernel of a group homomorphism exactly $K$. This proves both directions without requiring $\phi$ to be injective or surjective.

For the matrix application, reduce all entries modulo $p$:Reduction respects matrix multiplication and the determinant, so this is a group homomorphism. Its kernel of a group homomorphism consists exactly of matrices congruent to the identity modulo $p$, which are the given matrices. Therefore $K=\ker \rho$ is a normal subgroup, the level-$p$ principal congruence subgroup. Surjectivity of reduction is not needed for this argument.

Represent the Riemann sphere by the complex projective line, with finite $z$ represented by $[z:1]$ and infinity by $[1:0]$. An invertible matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts by $[u:v]\mapsto [au+bv:cu+dv]$, giving the stated fractional map on finite points. Applying $B$ and then $A$ is exactly matrix multiplication $AB$, including at poles and infinity. Thereforeso $\theta$ is a group homomorphism from the special linear group to the Möbius group.

Every Möbius transformation has an invertible complex matrix representative $A$. Choose $\rho \in \mathbb{C}$ with ${\rho }^2\det A=1$. Such a nonzero square root exists in $\mathbb{C}$, and $\rho A$ has determinant one while representing the same projective map. This proves surjectivity. If the map is the identity, fixing zero and infinity forces $b=c=0$, and fixing one forces $a=d$. The determinant-one condition gives $a^2=1$. Conversely both scalar matrices $I$ and $-I$ act trivially. HenceFor the normal forms, use the fact that a complex $2\times 2$ matrix has a Jordan normal form: either it is diagonalizable or it has one nontrivial Jordan block. For a determinant-one representative, distinct eigenvalues are $\lambda ,{\lambda }^{-1}$. Diagonalizing gives $\mathrm{diag}(\lambda ,{\lambda }^{-1})$, whose projective action is $z\mapsto {\lambda }^{2}z$. Distinctness means ${\lambda }^2\ne 1$, so $\mu ={\lambda }^2$ is neither zero nor one.

If the eigenvalue is repeated, the determinant condition gives $\lambda =\pm 1$. A diagonalizable representative would be $\lambda I$ and would give the identity, excluded here. A nontrivial Jordan block is similar to $\left(\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right)$, giving $z\mapsto z+1/\lambda =z\pm 1$. Any invertible conjugating matrix gives a Möbius transformation; rescaling it to determinant one does not change that map. Thus these matrix similarities are genuine conjugacies in the Möbius group, proving the requested Möbius conjugacy normal forms.

Conjugacy preserves fixed points bijectively. A nonidentity scaling has exactly the two fixed points zero and infinity, while either translation has exactly one, infinity. The identity fixes every point. This proves all three possibilities and excludes any further finite number of fixed points.

Finally, suppose $P\circ T\circ P^{-1}=S$, with $S(z)=z+\epsilon$ and $\epsilon =\pm 1$. Then $S^n(z)=z+n\epsilon \to \infty$ for every finite $z$ in the topology of the Riemann sphere; infinity remains infinity. Since $P^{-1}$ is continuous on that sphere,This is iteration of a one-fixed-point Möbius transformation. The limiting statement includes the unique fixed point itself and uses spherical convergence, which also handles intermediate iterates equal to infinity.

The identity belongs to the center of a group. If $a,b$ are central, then for every $g\in G$,so the center is a subgroup. For a central $a$, $ga{g}^{-1}=a$ for every $g$, giving $gZ(G)g^{-1}=Z(G)$. Thus the center is a normal subgroup.

Write a matrix of the Heisenberg group as $h(x,y,z)$. Direct matrix multiplication and inversion giveThe identity is $h(0,0,0)$ and each matrix has determinant one, so these formulas show closure under products and inverses inside $G{L}_3(\mathbb{R})$. Hence $H$ is a subgroup.

The two products $h(x,y,z)h(a,b,c)$ and $h(a,b,c)h(x,y,z)$ differ only in their upper-right entries, by $xc-az$. Commutation with every $h(a,b,c)$ therefore requires $xc=az$ for all real $a,c$. Setting $a=0,c=1$ forces $x=0$; setting $a=1,c=0$ forces $z=0$. These conditions are also sufficient. ThusDefine the surjective group homomorphism $F:H\to ({\mathbb{R}}^2,+)$ by $F(h(x,y,z))=(x,z)$. The multiplication formula gives $F(h{h}^{\prime })=F(h)+F(h^{\prime })$, and its kernel of a group homomorphism is precisely $Z(H)$. The first isomorphism theorem yieldsThis center quotient of the real Heisenberg group can also be checked directly: two elements have the same $(x,z)$ coordinates exactly when they differ by multiplication by a central element, and quotient multiplication adds these coordinates.

Write $W_n(i,j)$ for the number of locally admissible words $(x_0,\dots ,x_n)$ with $x_0=i$ and $x_n=j$. Here local admissibility means $A_{x_r{x}_{r+1}}=1$ for each consecutive pair; length counts symbols, so there are $n$ transitions. For $n=0$, the single-symbol word gives $W_0(i,j)={\delta }_{ij}=(A^0{)}_{ij}$, and for $n=1$ we have $W_1(i,j)=A_{ij}$.

To append the last symbol $j$, choose the penultimate symbol $k$. Every choice gives a locally admissible word exactly when its length-$n$ prefix is locally admissible and $A_{kj}=1$. Different prefixes or penultimate symbols give different words, and every admissible word is obtained once. ThereforeThis is precisely the rule for matrix multiplication. Mathematical induction consequently givesEquivalently, expanding the matrix entry sums the products $A_{i{x}_1}{A}_{x_1{x}_2}\cdots A_{x_{n-1}j}$ over all intermediate symbols; each product is the indicator function of one allowed transition chain.

The intended meaning of “allowed” here is local admissibility. If instead it means a word that actually occurs in some two-sided sequence in ${\Sigma }_A$, the assertion needs an additional extension assumption. For example, $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ permits the word $(1,2)$ and has $(A{)}_{12}=1$, but ${\Sigma }_A$ is empty because no symbol can continue after $2$. Having a nonzero entry in every row and every column is sufficient: one can successively choose a predecessor and a successor to extend every locally admissible word indefinitely in both directions. No such assumption is needed for the local counting formula.

For two upper triangular matrices and $i>j$,Indeed a potentially nonzero factor $A_{ik}$ requires $k\ge i$, while a potentially nonzero $B_{kj}$ requires $k\le j$. These requirements cannot both hold when $i>j$. Thus their matrix product is upper triangular.

Direct matrix multiplication for the given $A$ givesAdding $A^3+A^2-A$ yields $I$, so $A(A^2+A-I)=(A^2+A-I)A=I$. HenceEvery nonnegative matrix power is in $\mathrm{span}\{I,A,A^2\}$: reduce powers of degree at least three using $A^3=-A^2+A+I$. More formally, repeated reduction shows that products of any two polynomials of degree at most two in $A$ again lie in this span. Since $A^{-1}$ also lies in it, positive powers of $A^{-1}$ prove the same assertion for all negative integers. This is an instance of the integer powers for an annihilating polynomial with roots one and minus one, because $A^3+A^2-A-I=(A-I)(A+I{)}^2=0$.

To find the required entries for every integer, use the block upper-triangular structure. The top-left entry of $A^n$ is $1$ for positive and negative powers. Its lower-right block isThe identity $(I_2-N{)}^n=I_2-nN$ holds for all integers: multiplication adds the coefficients of $N$, and $(I_2-N{)}^{-1}=I_2+N$ proves the negative cases. ThereforeWrite $s=(-1{)}^n$ and $A^n={\alpha }_n{A}^2+{\beta }_{n}A+{\gamma }_{n}I$. Comparing those three entries givesSubtracting the first two equations finds ${\beta }_n$; the third then finds ${\alpha }_n$, and the first finds ${\gamma }_n$. The unique solution isFor clarity, substituting also gives the full matrix power

Suppose $x^2+y^2=0$ in the field $F$. If $y\ne 0$, division by $y^2$ gives $(x/y{)}^2=-1$, contrary to the hypothesis. Hence $y=0$ and then $x^2=0$; the absence of nonzero zero divisors in a field gives $x=0$. ThereforeFor the stated operations on $F^2$, addition is an abelian group operation: associativity and commutativity hold componentwise, its identity is $(0,0)$, and the negative of $(x,y)$ is $(-x,-y)$.

To check multiplication without omitting associativity, associate to each pair the matrixThis is an injection into the two-by-two matrices over $F$, and direct matrix multiplication givesIt also preserves addition. The associativity and distributivity of matrix multiplication therefore imply the same laws for the pair multiplication. Its formula is symmetric in the pairs, so it is commutative. Its identity is $(1,0)$, which is different from $(0,0)$.

For a nonzero pair, the first result shows that $x^2+y^2\ne 0$. Its multiplicative inverse issince multiplying $(x,y)$ by $(x,-y)$ gives $(x^2+y^2,0)$. This verifies every field axiom, so the given operations make $F^2$ a field. The embedded copy of $F$ is $\{(x,0):x\in F\}$; the element $(0,1)$ has square $(-1,0)$, describing a quadratic extension of $F$.

Preserving the Euclidean scalar product means $(Ax{)}^T(Ay)=x^{T}y$ for every pair of real vectors $x,y$. Since these bilinear expressions agree on all pairs of basis vectors, this is equivalent to $A^{T}A=I$: the columns of $A$ form an orthonormal basis.

The first column is a unit vector, so it is $(\cos \theta ,\sin \theta {)}^T$ for some real $\theta$. A perpendicular unit vector is one of the two choices $\epsilon (-\sin \theta ,\cos \theta {)}^T$, where $\epsilon \in \{1,-1\}$. Hence all the required orthogonal matrices areHere $R(\theta )$ is a planar rotation, and $\theta$ is determined modulo $2\pi$. To check the group laws explicitly, $D_{\epsilon }R(\varphi )=R(\epsilon \varphi )D_{\epsilon }$, soThis proves closure. The identity is $R(0)D_1=I$, the inverse of $R(\theta )D_{\epsilon }$ is $R(-\epsilon \theta )D_{\epsilon }$, and associativity is inherited from matrix multiplication. Thus the set is the orthogonal group $O(2)$.

For the Minkowski metric $J=\mathrm{diag}(1,-1)$, preservation means $(Ax{)}^{T}J(Ay)=x^{T}Jy$ for all real vectors $x,y$. This is equivalent to $A^{T}JA=J$, or, entrywise,The restriction $a>0$ puts the first column on the right branch of the unit hyperbola, so uniquely $a=\cosh t$, $c=\sinh t$ for a real parameter $t$. The third equation makes the second column a scalar multiple of $(\sinh t,\cosh t{)}^T$: this also covers $t=0$, when $b=0$. The second equation makes that multiple $\epsilon =\pm 1$. Thus all the required matrices areThe hyperbolic function addition formulas give $H(t)H(s)=H(t+s)$, while $D_{\epsilon }H(s)=H(\epsilon s)D_{\epsilon }$. ConsequentlyThe product remains of the required form and its upper-left entry is positive. The identity is $H(0)D_1$, and the inverse is $H(-\epsilon t)D_{\epsilon }$. Together with associativity of matrix multiplication, these identities establish the group property. This is the time-orientation-preserving Lorentz group in one spatial dimension.

The centre of a group is $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$. It is a normal subgroup. For $|G|=p^r$ with $r\ge 1$, let $G$ act on itself by conjugation. The orbit-stabilizer theorem makes the size of the conjugacy class of $g$ equal to $[G:C_G(g)]$, a power of $p$. A class has size one exactly when $g\in Z(G)$; every other class size is divisible by $p$. The class equation therefore givesso $p$ divides $|Z(G)|$. Since the identity belongs to the centre of a group, its size is a positive multiple of $p$. This proves the nontrivial center of a finite p-group.

If $G/Z(G)$ is cyclic, choose a representative $a$ of a generator. Every element is $a^{m}z$ for an integer $m$ and $z\in Z(G)$. Any two such elements commute, since their central factors commute with everything and their remaining factors are powers of the same element. Thus $G$ is an abelian group, $Z(G)=G$, and the cyclic quotient by the centre of a group is trivial.

For a non-abelian group of order $p^3$, the centre of a group has order $p$, $p^2$ or $p^3$ by the proved result and Lagrange's theorem. The last would make the group abelian. The middle would make $G/Z(G)$ have prime order and hence cyclic, also making $G$ abelian by the preceding argument. Therefore its centre has order $p$.

Finally write the given upper-unitriangular matrix as $M(a,b,c)$. Direct matrix multiplication givesIt commutes with every $M(a^{\prime },b^{\prime },c^{\prime })$ exactly when $a{c}^{\prime }=a^{\prime }c$ for all $a^{\prime },c^{\prime }\in F$. Taking $(a^{\prime },c^{\prime })=(0,1)$ forces $a=0$, and taking $(1,0)$ forces $c=0$; those conditions also suffice. Thus the centre of a group in this Heisenberg group over a prime field is

The matrix trace is the sum of the diagonal entries:Using the definitions of matrix multiplication and matrix transpose,This is the symmetry of the real Frobenius inner product.

By the definition of an eigenvalue, there is a nonzero eigenvector $v$ with $Av=\lambda v$. Induction using matrix multiplication givesstarting from $r=1$. Thus the same nonzero eigenvector establishes

At a fixed $t$, write $A=f(t)$ and $B=f^{\prime }(t)$. Differentiating the matrix exponential in the direction $B$ givesThe order of factors matters because matrix multiplication need not be commutative.

To prove the derivative of the matrix exponential formula, expand its power series. The coefficient linear in $h$ in $(A+hB{)}^k$ is ${\sum }_{j=0}^{k-1}{A}^{j}B{A}^{k-1-j}$. On bounded sets of matrices, the exponential series and its directional derivative series converge uniformly: the latter is dominated in operator norm by $\Vert B\Vert {\sum }_{k\ge 1}k\Vert A{\Vert }^{k-1}/k!$. Terms with two or more $hB$ factors have sum $O(h^2)$ locally. Hence the derivative isOn the other hand, the integral in the box expands absolutely asThe integral identity follows by repeated integration by parts. Grouping by $a+b=k-1$ gives the same derivative. Finally differentiability means $f(t+h)=A+hB+o(h)$; the locally bounded derivative of the exponential sends the $o(h)$ error to $o(h)$, proving the chain rule even when $f^{\prime }$ is not continuous. If $[f,f^{\prime }]=0$, the formula simplifies to $e^f{f}^{\prime }$.

Products below combine matrix multiplication with the wedge product of differential forms of differential forms. Differentiate the projection identity $p^2=p$ to getMultiplying by $p$ on both sides also gives $pdpp=0$; applying the same argument to $1-p$ gives $(1-p)dp(1-p)=0$. Thus at each point $dp$ is off-diagonal relative to $\mathrm{im}p\oplus \ker p$, whereas $(dp{)}^2$ is diagonal and commutes with $p$. For $\omega =p(dp{)}^2$,because $d(dp)=0$.

The graded Leibniz rule and cyclicity of the matrix trace give, since $\omega$ has even degree,Here $(d\omega ){\omega }^{k-1}$ is off-diagonal: $(dp{)}^3$ is off-diagonal and every power of $\omega$ is diagonal. Its matrix trace is zero. Hence $\boxed{d\mathrm{tr}({\omega }^k)=0}$ for every $k\ge 1$; for $k=0$ the trace is the constant matrix size.

For the determinant, work over the commutative algebra of even differential forms. If $f(0)=1$, write $f(\omega )=I+N$, where $N$ has strictly positive degree and is nilpotent for dimensional reasons. The finite power series identityholds over this algebra, as follows either from the formal determinant identity over characteristic-zero commutative rings or by differentiating $\det (I+sN)$ in the scalar variable $s$. Each $N^j$ is a linear combination of positive powers of $\omega$. The matrix trace of the logarithm is therefore a closed differential form, and its finite exponential is closed by the graded Leibniz rule. Thus $\boxed{d\det f(\omega )=0}$. The projection connection interprets $\omega$ as the curvature of $pd$ on $\mathrm{im}p$, but the proof above directly establishes both closedness claims.

For either coefficient system $\mathbb{Z}$ or ${\mathbb{F}}_p$, multiplying two matrices keeps all entries in that system, and multiplicativity of the determinant preserves determinant one. The identity matrix has determinant one, and for a determinant-one matrix the matrix inverse isIts entries again belong to the same coefficient system and its determinant is one. Together with associativity of matrix multiplication, these facts establish all the group axioms. Thus both sets are special linear groups.

Reduce each entry modulo five to defineReduction respects sums and products, so $\overline{MN}=\overline{M}\overline{N}$. It also respects the determinant, ensuring that the image has determinant one. Hence $\pi$ is a group homomorphism. Its kernel of a group homomorphism consists exactly of determinant-one integer matrices congruent to the identity modulo five:The determinant-one condition remains part of this set; the four integers cannot be chosen independently without it. A kernel of a group homomorphism is a normal subgroup, since $\pi (gk{g}^{-1})=\pi (g)e\pi (g{)}^{-1}=e$ for $k\in \ker \pi$. This is the principal congruence subgroup of level five.

The general linear group $G{L}_2({\mathbb{F}}_5)$ consists of invertible two-by-two matrices over the finite field ${\mathbb{F}}_5$, under matrix multiplication. Such a matrix has two linearly independent columns. There are $5^2-1=24$ choices for a nonzero first column, and $5^2-5=20$ choices for a second column outside its one-dimensional span. ThusThe determinant is a group homomorphismwhere ${\mathbb{F}}_5^{\times }=\{1,2,3,4\}$ is the multiplicative group of nonzero field elements. It is surjective because $\mathrm{diag}(t,1)$ has determinant $t$. Its kernel of a group homomorphism is $S{L}_2({\mathbb{F}}_5)$. The first isomorphism theorem gives a quotient of order four, whenceThe nonzero squares form the subgroup $\{1,4\}$ of ${\mathbb{F}}_5^{\times }$. Its inverse imageis a subgroup with index two. Indeed, composing the determinant with the quotient map to ${\mathbb{F}}_5^{\times }/\{1,4\}$ makes $H$ the kernel of a surjective group homomorphism to a group of order two. Thus $|H|=240$.

Write $U^{\ast }={\overline{U}}^T$ for the conjugate transpose. Taking determinants in $U^{\ast }AU=A$ gives $|\det U{|}^2\det A=\det A$. Since the Hermitian matrix $A$ is invertible, every such $U$ is invertible and $|\det U|=1$. The identity belongs to $U_A$; if $U,V\in U_A$, then $(UV{)}^{\ast }A(UV)=V^{\ast }AV=A$; and multiplying $U^{\ast }AU=A$ by $(U^{\ast }{)}^{-1}$ and $U^{-1}$ gives $(U^{-1}{)}^{\ast }A{U}^{-1}=A$. Associativity comes from matrix multiplication, so this is a group.

The Hermitian form $h_A(v,w)=v^{\ast }Aw$ obeys $h_A(Uv,Uw)=h_A(v,w)$ precisely when $U\in U_A$. Thus $U_A$ is the group of linear isometries of this nondegenerate Hermitian form; positivity of $A$ is unnecessary.

For $A=I$, $U$ is a unitary matrix. Here is a direct diagonalization of a matrix proof. Over $\mathbb{C}$, its characteristic polynomial has a root, giving a unit eigenvector $v$ with $Uv=\lambda v$. Preservation of the inner product gives $|\lambda |=1$. If $w\perp v$, then $⟨Uw,Uv⟩=⟨w,v⟩=0$, so $v^{\perp }$ is invariant. The restriction is again a unitary matrix. Induction on the dimension supplies an orthonormal basis of eigenvectors. In that basis, $U$ is diagonal, with diagonal entries on the unit circle.

Write the Heisenberg group element as $M(x,y,z)$. Matrix multiplication and inversion giveAll matrices have determinant one. They contain $M(0,0,0)=I$, are closed under multiplication and inverses, and inherit associativity from matrix multiplication; hence they form a subgroup of $G{L}_3(\mathbb{R})$.

Two such matrices commute precisely when $x{z}^{\prime }=x^{\prime }z$. Requiring this for every $x^{\prime },z^{\prime }$ forces $x=z=0$. ConsequentlyThe map $\pi (M(x,y,z))=(x,z)$ is a surjective group homomorphism to the additive group ${\mathbb{R}}^2$, with kernel of a group homomorphism $Z(H)$. Equivalently, it gives the explicit quotient group isomorphismIt is well-defined because multiplying by a central element changes only $y$. It is injective because equal $(x,z)$ imply that the quotient of the two matrices is central, and surjectivity follows by taking $y=0$. This also proves the needed instance of the first isomorphism theorem directly.

A counterclockwise planar rotation matrix has the formTo recover this form from preservation of the Euclidean norm, let $\mathbf{u},\mathbf{v}$ be the columns of $R$. Applying the norm condition to the two coordinate unit vectors gives $|\mathbf{u}|=|\mathbf{v}|=1$. Applying it to their sum givesso $\mathbf{u}\cdot \mathbf{v}=0$. The columns are an orthonormal basis, and $R^{T}R=I$: $R$ is an orthogonal matrix.

Write $\mathbf{u}=(\cos \theta ,\sin \theta {)}^T$. There are two perpendicular unit vectors available for $\mathbf{v}$, namely $\pm (-\sin \theta ,\cos \theta {)}^T$. The first gives determinant $1$, the second $-1$. The positive determinant hypothesis selects the first. A norm-preserving planar map with positive determinant is a rotation.

For the commutation condition, writeDirect matrix multiplication givesEquality forces $c=-b$ and $d=a$, so, with $u=a$ and $v=-b$,If $u=v=0$, take $\lambda =0$ and $R=I$. Otherwise set $\lambda =\sqrt{u^2+v^2}>0$ and choose $\cos \theta =u/\lambda$, $\sin \theta =v/\lambda$. ThenThis describes the centralizer of a planar quarter-turn. Its nonzero elements act as a rotation followed by a uniform scaling, just as multiplication by a nonzero complex number does.

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.