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The order of a group element $x$ is the least positive integer $r$ with $x^r=e$, or infinity if none exists. In a finite group, some two powers among $e,x,\dots ,x^{|G|}$ coincide; cancellation then gives a positive power equal to $e$, so $r$ exists. The generated cyclic group $H=⟨x⟩$ consists of the distinct elements $e,x,\dots ,x^{r-1}$ and has size $r$.

We now prove the needed divisibility directly by cosets. Multiplication by $g$ bijects $H$ with the left coset $gH$, so every left coset has $r$ elements. If $gH$ and $hH$ intersect, write $ga=hb$ with $a,b\in H$. Then $h^{-1}g=b{a}^{-1}\in H$, which implies $gH=hH$. Thus distinct left cosets are disjoint, and every element of $G$ belongs to its own left coset. The finite set $G$ is partitioned into, say, $k$ such cosets. Hence $|G|=kr$ andThis supplies the counting argument underlying Lagrange's theorem, rather than appealing to it or to an unproved orbit formula.

A divisor of $|G|$ need not occur as an element order. Use $G=C_2\times C_2\times C_2$, of order eight, and $d=4<8$. Every element has square equal to the identity, so there is no element of order four. Thus the proposed converse is false, even for an abelian group.

For the direct product of groups $C_m\times C_n$, let $a,b$ generate the factors. The least $k>0$ for which $(a,b{)}^k=(e,e)$ is the least common multiple of $m,n$. If $m,n$ are coprime, this is $mn$, the size of the whole product, so $(a,b)$ generates every element. Conversely, for any pair $(u,v)$ its order is $\mathrm{lcm}(\mathrm{ord}u,\mathrm{ord}v)$ and divides $\mathrm{lcm}(m,n)$. If $\gcd (m,n)>1$, that bound is $mn/\gcd (m,n)<mn$, so no element can generate the product. ThereforeThis is the cyclicity of a product of two finite cyclic groups; the general pair-order criterion is the order of a direct-product element.

Use suffix notation, with repeated indices summed from one to three. The cross product, divergence, and curl have components $(\mathbf{A}\times \mathbf{B}{)}_i={ϵ}_{ijk}{A}_j{B}_k$, $\nabla \cdot \mathbf{A}={\partial }_i{A}_i$, and $(\nabla \times \mathbf{A}{)}_i={ϵ}_{ijk}{\partial }_j{A}_k$. Here $ϵ$ is the Levi-Civita symbol.

The product rule givesUsing the cyclic identity ${ϵ}_{ijk}={ϵ}_{kij}$ in the first term and ${ϵ}_{ijk}=-{ϵ}_{jik}$ in the second identifies these terms as $B_k(\nabla \times \mathbf{A}{)}_k$ and $-A_j(\nabla \times \mathbf{B}{)}_j$. ThereforeFor the second identity, contract two Levi-Civita symbols, using ${ϵ}_{ijk}{ϵ}_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$:The four component expressions translate intoBoth derivations require only continuously differentiable vector fields; no additional divergence-free assumptions are used.

A group isomorphism $f:G\to H$ is a bijection satisfying $f(xy)=f(x)f(y)$ for all $x,y\in G$. It sends the identity to the identity and satisfies $f(x^k)=f(x{)}^k$. Applying its inverse shows that it preserves the order of a group element exactly.

The cyclic group $C_8$ has an element of order eight. Every element of the direct product of groups $C_4\times C_2$ has fourth power equal to the identity, and a generator of the first factor paired with the identity of the second has order four. Every nonidentity element of $C_2\times C_2\times C_2$ has order two. Thus their maximum element orders are respectively $8,4,2$, which an isomorphism would preserve. The three groups are pairwise nonisomorphic.

For a fourth example take the dihedral group of symmetries of a square. It consists of four rotations and four reflections, so has order eight. With $r$ a quarter-turn and $s$ a reflection, $srs=r^{-1}\ne r$, showing $sr\ne rs$. This group is nonabelian, whereas all three displayed direct products of groups are abelian groups. Since a group isomorphism preserves commutativity, the square's dihedral group is isomorphic to none of them. Its order is eight regardless of whether a notation convention calls it $D_4$ or $D_8$.

Let $p$ be a permutation of the finite set. Starting from $a$, repeatedly apply $p$. Some iterate repeats; because $p$ is invertible, the first repeat closes the sequence at $a$. More explicitly, if $p^i(a)=p^j(a)$ with $0\le i<j$, applying $p^{-i}$ gives $p^{j-i}(a)=a$. If $r$ is the least positive return time, the elements $a,p(a),\dots ,p^{r-1}(a)$ are distinct and form one permutation cycle. If the set is not exhausted, start again at an unused element. Two such sets cannot overlap: an overlap lets an inverse iterate express either starting point as an iterate of the other, making the sets identical. Thus this process partitions the set into disjoint permutation cycles. On each part its cycle acts exactly as $p$ and the other cycles fix the elements. Their product therefore equals $p$. This proves existence of the cycle decomposition of a permutation, including one-cycles for fixed points.

Use the composition convention that the rightmost permutation acts first. For the specified product, the images areHenceThe order of a group element here is the least positive exponent making every cycle return to the identity. It must be divisible by each cycle length, and any common multiple does make every cycle return. The least common multiple of $3,2,3$ is six.

For a Cartesian second-rank tensor, writeThe first term is a symmetric second-rank tensor and the second is an antisymmetric second-rank tensor. Taking the trace of the symmetric term and subtracting it gives the scalar, symmetric-traceless and axial tensor decomposition:Here $S_{ij}=S_{ji}$ and $S_{ii}=0$. To check reconstruction of the antisymmetric part, contract the Levi-Civita symbols:Thus $P_{ij}=P{\delta }_{ij}+S_{ij}+{ϵ}_{ijk}{A}_k$, and the displayed contractions also prove uniqueness. The vector associated with the antisymmetric part is an axial vector under reflections, and an ordinary vector under proper rotations.

For the specified isotropic tensor law, contract the Kronecker deltas first:Write $T_{ij}=T{\delta }_{ij}+W_{ij}+{ϵ}_{ijk}{V}_k$, where $T=T_{kk}/3$ and $W$ is symmetric and traceless. Transposition preserves its scalar and symmetric-traceless parts and reverses its antisymmetric part. ThereforeUniqueness of the decomposition gives the isotropic elasticity on tensor components relations. Provided their respective coefficients are nonzero, the requested inverses areIf one coefficient vanishes, that strain component has zero stress response: the corresponding stress component must be zero for a solution to exist, and that strain component is not uniquely recoverable. Isotropy by itself does not guarantee invertibility.

If $T_{ij}$ is symmetric then $V=0$, hence $A=0$ and the stress is symmetric. Equivalently, direct transposition gives $P_{ij}-P_{ji}=(\beta -\gamma )(T_{ij}-T_{ji})=0$. Restricting to this symmetric-strain case givesThis final two-parameter expression uses the symmetry assumption from the preceding request. For a general nonsymmetric array, the transpose term cannot be absorbed into the same coefficient: it acts with $\beta +\gamma$ on the symmetric-traceless part and with $\beta -\gamma$ on the antisymmetric part. The stated coefficient $\mu$ is the coefficient of $T_{ij}$ itself; if one instead writes the conventional elastic law with $2\mu$ multiplying the strain, that shear-modulus convention is half this coefficient.

The divergence of a cross product identity and the stated static field equations giveFor an outward-oriented closed surface, the divergence theorem therefore gives the flux of the specified Poynting vector normalization:No open-surface conclusion follows from this equation without including its closing pieces.

For the particular fields, direct calculation givesThe fields also have $\nabla \times \mathbf{E}=0$, $\nabla \cdot \mathbf{B}=0$, and $\nabla \cdot \mathbf{J}=0$, consistently with the required equations. Close the outward spherical patch by the two planar half-disks on $x=0$ and $y=0$, forming the quarter unit ball $V$. Its total closed flux iswhere $r$ in this integral is cylindrical radius. The $x=0$ face has outward normal $-{\mathbf{e}}_x$ and zero flux because $P_x=0$ there. The $y=0$ face has outward normal $-{\mathbf{e}}_y$, so its flux isSubtracting these planar contributions from the closed flux gives the requested open-patch answerAs a direct check, on the unit sphere the outward normal is $(x,y,z)$ and $\mathbf{P}\cdot \mathbf{n}=xy(x^2-2z^2)$. With spherical angles $0\le \vartheta \le \pi$, $0\le \phi \le \pi /2$, its surface integral is $\frac{1}{4}{\int }_0^{\pi }{\sin }^5\vartheta d\vartheta -{\int }_0^{\pi }{\sin }^3\vartheta {\cos }^2\vartheta d\vartheta =\frac{1}{4}(16/15)-4/15=0$. The curved patch's vanishing flux is a cancellation, not an inference that its flux density vanishes everywhere.

For a nonnegative integer-valued random variable, its expected value is $\mathbb{E}N={\sum }_{r=0}^{\infty }r{p}_r$, allowing the value $+\infty$. Since $r={\sum }_{n=1}^{r}1$, rearranging nonnegative summands gives the tail-sum formula for expectation:No integrability assumption is needed for the rearrangement. The same formula holds for an extended nonnegative integer-valued variable by the monotone convergence theorem, using $N={\sum }_{n\ge 1}{1}_{\{N\ge n\}}$.

For the first ascent in independent continuous observations, continuity makes ties probability zero. By order symmetry of independent random variables, the first $m$ observations have each of their $m!$ strict orderings with equal probability. For $n\ge 2$, $N\ge n$ means exactly that the first $n-1$ observations are decreasing, soSubtracting consecutive tails giveswith probability zero for $r=0,1$. The events $\{N\ge n\}$ decrease to $\{N=\infty \}$; continuity of probability under decreasing intersections gives $P(N=\infty )={\lim }_{n}1/(n-1)!=0$. Finally,The first observation cannot itself end the run; this is why the tail includes the separate term $P(N\ge 1)=1$.

Write $u(x,y)=U(\xi ,\eta )$. The chain rule gives ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ and ${\partial }_y=2{\partial }_{\xi }+3{\partial }_{\eta }$. Consequently the coefficients of $U_{\xi \xi }$ and $U_{\eta \eta }$ cancel, while that of $U_{\xi \eta }$ is $12-25+12=-1$. The transformed partial differential equation is $-U_{\xi \eta }=1$. Integrating once in each variable gives the most general twice differentiable solution,where $F,G$ are arbitrary twice differentiable functions. The change of variables is invertible because its Jacobian determinant is $3-2=1$.

On $y=0$, the value condition gives $F(x)+G(x)=x^2$. Differentiating it yields $F^{\prime }(x)+G^{\prime }(x)=2x$. The derivative condition gives $-5x+2F^{\prime }(x)+3G^{\prime }(x)=x$. Subtraction therefore gives $G^{\prime }(x)=2x$ and $F^{\prime }(x)=0$. The constants in $F$ and $G$ cancel in their sum, leavingIndeed $u_{xx}=0$, $u_{xy}=1$, $u_{yy}=6$, so the original operator gives $-5+6=1$, and the two initial-line conditions hold.