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The orbit of $(1,\dots ,k)$ consists precisely of the ordered $k$-tuples of distinct elements of $\{1,\dots ,n\}$:It has sizeThe stabilizer consists of permutations fixing $1,\dots ,k$ pointwise, while freely permuting the remaining letters:Thusverifying orbit-stabilizer.

For $x\in X$,The stabilizer contains the identity; if $g,h$ fix $x$, thenso it is a subgroup.

The orbit-stabilizer theorem statesDefineThe map is well defined and bijective: two elements give the same image exactly when they differ by an element of the stabilizer. Lagrange's theorem then proves the formula.

No. The alternating group $A_4$ has order twelve, and six is a proper divisor of twelve, but $A_4$ has no element of order six. Its nonidentity elements are eight 3-cycles of order three and three double transpositions of order two.

The order of an element $x$ is the least positive integer $d$ such that $x^d=e$, and the order of a finite group is its number of elements.

Lagrange theorem states that if $H\le G$ and $G$ is finite, thenIndeed, the left cosets of $H$ partition $G$. Multiplication by a coset representative is a bijection from $H$ to each coset, so every coset has $|H|$ elements. Summing over the $[G:H]$ cosets proves the formula.

The cyclic subgroup $⟨x⟩$ has exactly $d$ elements when $x$ has order $d$. Applying Lagrange to $⟨x⟩\le G$ gives

For $F=(y,z,x)$,Its antisymmetric part isAlso,For example, $-(1/2){ϵ}_{12k}{B}_k=1/2$, and the same componentwise calculation gives

A second-rank tensor is antisymmetric whenUnder an orthogonal Cartesian coordinate change with matrix $R$,Thereforeafter relabelling the dummy indices. Hence the antisymmetric second-rank tensor property is independent of Cartesian coordinates.

In Einstein notation,This is

Similarly,after cyclically relabelling the first term and swapping two indices in the second. Hence

Such a group does exist. Take $H=S_3$. Its center is trivial, sohas order six. Therefore

Suppose $G/Z(G)$ is generated by $gZ(G)$. Every element of $G$ can then be written $g^{a}z$ with $z\in Z(G)$. HenceThus the cyclic quotient by the center condition forces $G$ to be abelian.

If $|H/Z(H)|=7$, the quotient has prime order and is therefore cyclic. The result just proved would make $H$ abelian, so $Z(H)=H$ and the quotient would have order one, a contradiction. Therefore

Both divergences vanish, so $\rho =0$. Faraday's and Ampere's curl equations respectively requireFor positive constants these imply

The Poynting vector isOnly the two $x$-faces of the box contribute to its outward flux. With $a=\pi /(2k)$,The electromagnetic energy inside the box isand direct differentiation, using $\omega =k$, givesThusconfirming the integral identity when $J=0$.

The vector identity from Question 3 givesIntegrating over $V$ and applying the divergence theorem proves the Poynting theorem

For $u=(-y,x,\alpha )$, one has $\nabla \cdot u=0$, and therefore

Close the upper half-ellipsoid with its unit-disk base $D$ in the plane $z=0$. By symmetry,On the base, the outward normal is $n=(0,0,-1)$, soThe first two components integrate to zero over the disk, while the third integrates to $-\pi {\alpha }^2$. The closed-surface tensor identity therefore givesMultiplying by ${\beta }^2$,

Fix the first index $i$ and form the vector field with components $F_j=T_{ij}$. The divergence theorem givesSubstitution yields the tensor divergence theorem

SetThe ellipsoid becomes the unit ball and the Jacobian determinant is $abc$. Thereforeassuming $a,b,c>0$.