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Let $\rho =3m/(4\pi R_0^3)$ be the uniform mass density. In spherical coordinates $(r,\vartheta ,\varphi )$ with the chosen diameter as polar axis, the squared perpendicular distance to that axis is $r^2{\sin }^2\vartheta$. The moment of inertia is thereforeThis derives the moment of inertia of a uniform solid sphere about any central axis by spherical symmetry.

Put $a=R_1-R_0$. The center of mass follows the circle of radius $a$, with position ${\mathbf{r}}_C=a(\sin \theta ,-\cos \theta )$. Let ${\mathbf{e}}_r=(\sin \theta ,-\cos \theta )$ and ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$. Its velocity is $a\dot{\theta }{\mathbf{e}}_{\theta }$, and the contact point is displaced from the center of mass by $R_0{\mathbf{e}}_r$. With positive spin angular velocity ${\Omega }_s$ counterclockwise, its contact-point velocity isThe cylinder is fixed, so rolling without slipping requires this velocity to vanish. Consequently ${\Omega }_s=-a\dot{\theta }/R_0$. If instead the spin angle $\psi$ is positive clockwise, opposite to increasing $\theta$, then the printed positive-coefficient convention isThe spin and orbital rotation have opposite directions when both are measured about the same oriented axis; the angular speed is $a|\dot{\theta }|/R_0$. This specifies the sign convention in the printed expression.

The instantaneous contact point is at rest, so the normal reaction and static friction do no work. Conservation of energy equates the loss of gravitational potential energy from release to the total kinetic energy:On the descent from $0<\alpha <\pi /2$, $\theta$ decreases, soIntegrating $dt=-d\theta /|\dot{\theta }|$ from release to the bottom givesThe integral is finite: at the release point its denominator is asymptotic to $\sqrt{\sin \alpha (\alpha -\theta )}$. Contact is maintained, since the inward radial equation of motion gives the positive normal reaction $N=m(g\cos \theta +a{\dot{\theta }}^2)$ throughout this range.

For frictionless sliding, the normal reaction passes through the sphere's centre and produces no torque; Newtonian gravity also exerts no central torque. With initial zero spin, the sphere remains nonrotating. All the lost gravitational potential energy becomes translational kinetic energy, soA hollow spherical shell of the same mass and radius places its material farther from the centre, giving a larger moment of inertia. Under rolling without slipping, a given centre speed therefore requires more rotational kinetic energy; the same gravitational drop yields a smaller speed. Replacing the solid sphere by a hollow shell increases $T_R$ and leaves $T_S$ unchanged. The sliding case remains nonrotating and has the same center of mass path and gravitational drop. This is the general comparison for rolling inside a fixed circular cylinder.

Let a finite group $G$ act on a set $X$. For $x\in X$, define its stabilizer subgroup $G_x=\{g:g\cdot x=x\}$ and its group orbit $Gx=\{g\cdot x:g\in G\}$. The mapis well-defined, because multiplying $g$ on the right by an element fixing $x$ does not change its image. It is onto by the definition of the orbit. If $g\cdot x=h\cdot x$, then $h^{-1}g\in G_x$, so $g{G}_x=h{G}_x$; hence it is one-to-one. Every coset has $|G_x|$ elements and the cosets partition $G$. This proves the orbit-stabilizer theorem:For a group element $x$, its cyclic subgroup $H=⟨x⟩$ has $\mathrm{ord}(x)$ elements. Apply the theorem to the action of $G$ on its left cosets $G/H$, whose stabilizer at $H$ is $H$. It follows thatThis is the needed case of Lagrange's theorem, obtained directly from the action.

To prove Cauchy's theorem for finite groups, let $p$ be a prime divisor of $|G|$ and considerThe first $p-1$ entries are arbitrary and determine the last, so $|\mathcal{X}|=|G{|}^{p-1}$ is divisible by $p$. Cyclically rotate the entries. This preserves the product constraint even if $G$ is nonabelian, sinceThus a cyclic group of order $p$ acts on $\mathcal{X}$. By the orbit-stabilizer theorem, its orbits have size one or $p$. Its fixed tuples are exactly $(g,\dots ,g)$ with $g^p=e$. If $m$ is their number, counting the remaining size-$p$ orbits gives $m\equiv |\mathcal{X}|\equiv 0(\mathrm{mod}p)$. The identity provides one fixed tuple, so $m\ge p$ and a nonidentity $g$ with $g^p=e$ exists. The order of this $g$ divides $p$: division of $p$ by its order and minimality of that order prove this directly. Since $p$ is prime and $g\ne e$, the order is exactly $p$. This is the cyclic-tuple proof of Cauchy theorem.

Now suppose every nonidentity element has order two. Then every element equals its inverse, and for any $a,b\in G$,Thus group of exponent two is abelian applies. Starting from the identity subgroup, choose an element outside the subgroup already generated. Because the group is abelian and that element squares to the identity, adjoining it doubles the subgroup: it is the disjoint union of the old subgroup and its translate by the new element. Repeating until the finite group is exhausted shows $|G|=2^r$. Conversely the direct product of groups $C_2^r$ has order $2^r$ and every nonidentity element has order two. ThereforeThe case $r=0$ is the trivial group and satisfies the condition vacuously. These finite examples are elementary abelian groups of exponent two.

An infinite example is the group of all binary sequences, with coordinatewise addition modulo two:Every sequence added to itself is zero, so every nonzero sequence has order exactly two. The sequences having a single $1$ in coordinate $j$ are all distinct, proving that the group is infinite.

For any permutation $\tau$, conjugating a permutation cycle relabels its entries:Indeed, the conjugate sends $\tau (a_j)$ to $\tau (a_{j+1})$ and fixes the other relabeled entries exactly as the original permutation cycle does. Therefore conjugation preserves the multiset of permutation cycle lengths. Conversely, if two permutations have the same cycle type, pair their permutation cycles of each length and let $\tau$ carry the $j$th entry of each first permutation cycle to the $j$th entry of its partner. Include one-cycles for fixed points. This defines a bijection of all labels, and the displayed relabeling identity makes the permutations conjugate. ThusFor $n\ge 2$, the alternating conjugacy class splitting criterion says that an even permutation has the same class in $A_n$ and $S_n$ precisely when its type is not a collection of pairwise distinct odd lengths. Equivalently, there is an even-length permutation cycle or a repeated permutation cycle length, counting each fixed point as a length-one permutation cycle. Distinct odd lengths instead give two alternating-group classes. For $n=1$, $A_1=S_1$ and the sole class is the same; the index-two splitting criterion is not applicable.

Squaring an odd-length permutation cycle keeps one permutation cycle of that length, since advancing by two visits all of its entries. Squaring an even-length permutation cycle splits it into two permutation cycles, each of half that length. Thus any even-length permutation cycle strictly increases the total permutation cycle count on squaring, while odd permutation cycles leave that count unchanged. The cycle type can agree with its square exactly when every length is odd. This proves permutation conjugate to its square:The final equivalence follows because the order of a group element that is a permutation is the least common multiple of its permutation cycle lengths.

The possible even types on five labels are the identity, a three-cycle, a product of two disjoint transpositions and a five-cycle. The identity and the double transpositions are their own inverses. For a three-cycle $(abc)$ with unused labels $d,e$, the even permutation $(bc)(de)$ conjugates it to its inverse. For a five-cycle $(abcde)$, the even permutation $(be)(cd)$ reverses its order. Therefore every element of $A_5$ is conjugate to its inverse within $A_5$. Inverting a five-cycle uses two transpositions, in agreement with inversion of an odd cycle in an alternating group.

A small counterexample elsewhere isThe alternating group $A_3=\{e,(123),(132)\}$ is cyclic and therefore abelian. Every conjugacy class in an abelian group is a singleton, while $\sigma \ne {\sigma }^{-1}$. Thus these two elements are not conjugate in $A_3$.

For a constant orthogonal Cartesian coordinate change $x_i^{\prime }=R_{ij}{x}_j$, a Cartesian second-rank tensor has components satisfyingBoth indices transform by the coordinate-change matrix; summation over repeated indices is understood by the Einstein summation convention. A vector would have the one-index rule $U_i^{\prime }=R_{ik}{U}_k$.

Define $U_i=\partial T_{ij}/\partial x_j$. Since $x_m=R_{jm}{x}_j^{\prime }$, the chain rule gives $\partial /\partial x_j^{\prime }=R_{jm}\partial /\partial x_m$. The rotation coefficients are constant, soThe identity $R_{jm}{R}_{jl}={\delta }_{ml}$ follows from orthogonality. Hence the contracted derivative transforms as a vector. This proves the divergence of a Cartesian second-rank tensor rule for global Cartesian rotations. In curvilinear coordinates the corresponding construction uses a covariant derivative; ordinary partial derivatives alone would not have the stated tensorial transformation law.

The real orthogonal group and special orthogonal group are, under matrix multiplication,An orthogonal matrix preserves Euclidean inner products; taking determinants of $Q^{T}Q=I$ shows that its determinant is $1$ or $-1$. Its inverse is $Q^T$, so $Q{Q}^T=I$ as well.

For $Q\in SO(3)$, use $Q-I=Q(I-Q^T)$. The determinant then satisfiesIt follows that $\det (Q-I)=0$. Thus the real matrix $Q-I$ has a nonzero vector in its kernel of a linear map, yieldingThis is the three-dimensional case of odd-dimensional special orthogonal transformation has a fixed vector; geometrically it supplies the fixed rotation axis, unless the transformation is the identity and every direction is fixed.

The claim is false for $O(3)$. The matrix $Q=-I_3$ is orthogonal with determinant $-1$ and has only the eigenvalue $-1$. It therefore has no nonzero eigenvector with eigenvalue $1$.

For a permutation $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix, defined by $P_{\sigma }{e}_j=e_{\sigma (j)}$. Define the signature of a permutation by $ϵ(\sigma )=\det P_{\sigma }$. A permutation matrix is orthogonal, so $(\det P_{\sigma }{)}^2=1$ and its permutation sign lies in $\{1,-1\}$. A transposition exchanges two columns of the identity and therefore has determinant $-1$. Every permutation is a product of transpositions, by decomposing each permutation cycle asHence $ϵ(\sigma )=(-1{)}^m$ for any expression as $m$ transpositions; the determinant definition proves that this parity of a permutation is independent of the expression. Equivalently the permutation sign is positive for an even permutation and negative for an odd permutation.

Composition is taken rightmost first. Since $P_{\sigma \tau }{e}_j=e_{\sigma (\tau (j))}=P_{\sigma }{P}_{\tau }{e}_j$, we have $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$. Multiplicativity of the determinant therefore proves the sign homomorphism identityThe alternating group consists of the even permutations:It contains the identity. If $\sigma ,\tau \in A_n$, their product has permutation sign $1$, and $ϵ({\sigma }^{-1})=ϵ(\sigma {)}^{-1}=1$, so their inverses also belong to $A_n$. This proves it is a subgroup. Finally, for every $g\in S_n$ and $a\in A_n$,Thus $A_n$ is a normal subgroup of $S_n$. This also exhibits directly why a kernel of a group homomorphism is normal.