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If A and B are commuting self-adjoint operators with $A^2=B^2=I$, then the displayed combination is a unitary operator. Indeed,and the reverse product is also I. The factors are involutions, and their commutation is essential to cancel the cross terms. Operators $A=U_A\otimes I$ and $B=I\otimes U_B$ on separate tensor factors provide a natural example. More generally, $\cos \theta A+i\sin \theta B$ is unitary for real $\theta$ by the same proof. Replacing i by a real coefficient usually loses this property.

A C-star isomorphism is a bijective C-star homomorphism: a complex-linear multiplicative map preserving the involution. Between unital C-star algebras it automatically preserves the unit. It preserves the Banach algebra spectrum, since it preserves invertibility in both directions. Applying the C-star identity and the spectral norm formula to $a^{\ast }a$ gives $\Vert \Phi (a){\Vert }^2=\Vert \Phi (a^{\ast }a)\Vert =\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$, so it is automatically isometric.

Every finite group of odd order is a solvable group. In particular, a nonabelian simple group has even order and therefore has an involution by Cauchy's theorem for finite groups. This is a deep theorem, not an elementary consequence of the Sylow theorems.

Let $W_g$ be a connected sum of oriented manifolds consisting of $g$ copies of $S^r\times S^r$ with odd $r\ge 1$. Suppose an orientation-preserving smooth involution $f$ has finitely many fixed points, all nondegenerate with positive determinant $\det (I-D{f}_x)$. The Lefschetz-Hopf fixed-point theorem and the cohomology ring of a connected sum of odd-dimensional sphere products giveThis middle group has dimension $2g$ and its Poincare duality pairing is alternating. The induced map preserves the pairing and squares to the identity. By eigenspaces of a symplectic involution, its trace is $2g-4b$ for an integer $b$. Therefore $\#\mathrm{Fix}(f)\equiv 2-2g(\mathrm{mod}4)$.

Double a compact orientable genus-zero topological surface with $g+1$ boundary circles. The resulting closed surface has genus $g$. The involution exchanging the halves fixes the seam circles and has local collar form $(s,t)\mapsto (s,-t)$. Compose it with a flow rotating each seam coordinate by a small nonzero amount, with speed supported in its collar. Equality with the initial point would force $t=0$, where the rotation is nonzero; off the collars the original reflection already has no fixed points. The new map is fixed-point-free and homotopic to the reflection through the flow. This gives a surface example in every genus where having fixed points is not a homotopy invariant.

On the nonzero real numbers, sign reversal $a$ and reciprocation $b$ are commuting involutions. Their function composition generates exactly $e(x)=x$, $a(x)=-x$, $b(x)=1/x$ and $ab(x)=-1/x$. The relations $a^2=b^2=e$ and $ab=ba$ identify the group with $C_2\times C_2$, the Klein four-group. Its subgroups are the trivial group, its three order-two cyclic subgroups and itself. Excluding zero from the domain is necessary for all four transformations to be defined and invertible.

Up to group isomorphism, the groups of order eight are $C_8$, $C_4\times C_2$, $C_2^3$, $D_8$ and $Q_8$, with $D_8$ here denoting the eight-element dihedral group. An element of order eight gives the first case. If every nonidentity element has order two the group is an elementary abelian group. Otherwise, after excluding elements of order eight, a cyclic subgroup of order four has index two; an element outside it either commutes with its generator or inverts it. In the latter case its square is either the identity or the central involution, giving the dihedral group or quaternion group.

Up to group isomorphism, the finite groups with at most six elements are the trivial group, the cyclic groups $C_2,C_3,C_4,C_5,C_6$, the Klein four-group $C_2\times C_2$, and the symmetric group $S_3$.

A group of prime cardinality is a cyclic group, since Lagrange's theorem forces any nonidentity element to generate it. For four elements, either an element has order of a group element four, or every nonidentity element is an involution. In the latter case the group of exponent two is abelian identity shows that two distinct nonidentity elements $a,b$ generate $\{e,a,b,ab\}\cong C_2\times C_2$.

For a group of order six, either an element generates it or some element $a$ has order of a group element three: otherwise all nonidentity elements would be involutions, producing a four-element subgroup in contradiction with Lagrange's theorem. Pairing each element with its inverse element gives an involution $b$ outside $⟨a⟩$. The subgroup $⟨a⟩$ has index of a subgroup two and is a normal subgroup, so $ba{b}^{-1}=a$ or $a^{-1}$. The commuting case makes $ab$ have order of a group element six, whereas the second case gives the multiplication rules of $S_3$ on the six distinct elements $a^i,a^{i}b$, $0\le i<3$.

All these groups embed in $S_4$ except $C_5$ and $C_6$. The former is excluded by Lagrange's theorem; the latter is excluded because the disjoint cycle decomposition of a permutation of four letters cannot have least common multiple of cycle lengths equal to six.

For a group $G$, inversion sends the conjugacy class of $g$ to that of $g^{-1}$, because $(xg{x}^{-1}{)}^{-1}=x{g}^{-1}{x}^{-1}$. It therefore defines an involution on the collection of conjugacy classes, even in a noncommutative group where inversion is not a group homomorphism.

A nonidentity involution in the affine group over a finite field ${\mathbb{F}}_{\ell }$, for odd prime $\ell$, has linear coefficient $-1$: a translation has order $\ell$, while $u^2=1$ forces $u=\pm 1$. Its unique fixed point is $b/2$, and its remaining orbits have size two. This orbit count computes primes in an intermediate field as double cosets for an order-two decomposition group acting on the radical roots.

// Target: number-theory.bigb

For the dihedral group of a regular $2n$-gon, $n\ge 2$, use $r^{2n}=s^2=1$ and $srs=r^{-1}$. Its involutions are the central half-turn $r^n$ and all reflections $s{r}^j$. The half-turn has singleton conjugacy class and normal closure $\{1,r^n\}$. Conjugating a reflection changes its exponent by an even integer or negates it, so its conjugacy class is $\{s{r}^{j+2k}:0\le k<n\}$. Products of consecutive reflections generate $r^2$, giving normal closure $⟨r^2,s{r}^j⟩$, of order $2n$ and index two. The two reflection classes are distinguished by exponent parity.

An involution partitions its set into fixed points and pairs exchanged by the function. If there is exactly one fixed point, every finite model has odd size. Conversely any finite set of odd size admits such a function by pairing all but one element. The first-order sentence $\forall xf(f(x))=x\wedge \exists z(f(z)=z\wedge \forall x(f(x)=x\to x=z))$ consequently has exactly the positive odd finite model sizes. Its arbitrarily large finite models imply an infinite model by the compactness theorem.

Reverse the order of a finite list. Reversal is an involution and satisfies $(AB{)}^R=B^R{A}^R$ for list concatenation. In an array segment of length $n$, exchanging symmetric offsets $j$ and $n-1-j$ for $0\le j<\lfloor n/2\rfloor$ realizes reversal in constant auxiliary space.

Suppose an involution $R$ of a Markov chain satisfies $P(Rx,Ry)=P(x,y)$. Start one chain at $x$, set the other equal to its image under $R$ until the first hitting time of the fixed-point set of $R$, and subsequently use identical transitions. This is a coalescing coupling, because the symmetry preserves each marginal transition matrix and both chains coincide at that hitting time.

Every nontrivial quotient group of a group generated by two involutions is generated by two involutions, allowing repeated generators. The two generator images have squares equal to the identity. If one image becomes trivial, the other generates $C_2$; if both become trivial, the quotient is trivial and is excluded. Thus the two-involution characterization of a dihedral group is preserved by nontrivial quotients.

An odd involution $A$ reverses fermion parity and is its own inverse, so it identifies the bosonic and fermionic subspaces. They have equal dimension in a finite-dimensional representation. For a supercharge $q$ with $q^2=0$ and $\{q,q^{†}\}=cI$, $c>0$, use $A=(q+q^{†})/\sqrt{c}$. This gives a direct proof of boson-fermion degeneracy in a supermultiplet at positive energy. A zero-energy supersymmetric vacuum need not have an accompanying fermionic vacuum.

Suppose real symmetric matrices $E,F$ satisfy $E^2=F^2=I$ and $EF=-FE$. The involutions are orthogonal matrices, and $J=EF$ satisfies $J^T=-J$, $J^2=-I$. The matrix exponential therefore reduces to $R(t)=e^{tJ}=\cos tI+\sin tJ$. Multiplication gives $R(t{)}^{T}R(t)=I$ and the conjugation identitiesIndeed $JE=-F$, $EJ=F$, $JF=E$, and $FJ=-E$, which give both identities by expansion. The angle in this two-dimensional space of matrices is twice the parameter in the orthogonal matrix. These formulas hold in any finite dimension where the hypotheses can be satisfied.

The quaternion group of order eight has a unique involution, which generates its center of a group. A nontrivial intersection $S\cap S^{\prime }$ is a nontrivial finite $2$-group, so by Cauchy's theorem for finite groups it contains an involution. This must be both the unique involution $t$ of $S$ and the unique involution of $S^{\prime }$. Thus $t\in Z(S^{\prime })$.

It follows that $S^{\prime }\le C_G(t)=N_G(S)$. But $S$ is a normal Sylow subgroup of its normalizer, so it is that normalizer's unique Sylow subgroup. Hence $\boxed{S^{\prime }=S}$. Distinct Sylow subgroups therefore intersect trivially. From the preceding result,Of $1,3,7,21$, only $1$ is congruent to $1$ modulo $8$. Therefore $n_2=1$, so $S$ is normal in $G$.

All 63 involutions lie in Sylow subgroups for $2$. Each such subgroup has exactly seven nonidentity elements and distinct ones have no nonidentity overlap. Hence