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The orbit-stabilizer theorem states that for a finite group $G$ acting on a set $X$ and $x\in X$,Define $G/G_x\to Gx$ by $g{G}_x\mapsto gx$. This is well-defined because $g{G}_x=h{G}_x$ exactly when $h^{-1}g\in G_x$, equivalently $gx=hx$. The same equivalence proves injectivity, and the definition of the orbit proves surjectivity. Thus $|Gx|=[G:G_x]$, giving the formula.

Because $n$ is not a prime power, distinct primes $p<q$ divide $n$. By Cauchy's theorem, $G$ has subgroups $H,K$ of orders $p,q$. Combine the two coset actions to obtainThe kernel of the first action is contained in $H$, and that of the second is contained in $K$. Their intersection is trivial because $H\cap K=\{1\}$, so the combined action is faithful.

Since $p\ge 2$, $q\ge 3$, and $n\ge 6$,Adding fixed points embeds this symmetric group into $S_{n-1}$. Thus every group of non-prime-power order $n$ is a subgroup of $S_{n-1}$, as summarized by symmetric-group embedding at one less than the group order.

Let $n=p^k$ and take $G=C_{p^k}$. If it embedded in $S_{n-1}$, that symmetric group would contain a permutation of order $p^k$. The order of a permutation is the least common multiple of its cycle lengths. For this least common multiple to be divisible by $p^k$, one cycle length must itself be divisible by $p^k$, requiring at least $p^k=n$ points. This is impossible in $S_{n-1}$. Thus $C_n$ is the required group.

The Cauchy theorem for groups says that if a prime $p$ divides $|G|$, then $G$ contains an element of order $p$.

ConsiderThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$, which is divisible by $p$. Cyclic rotation acts on $X$: if the product is one, thenEvery orbit has size one or $p$. The fixed points are exactly the constant tuples $(x,\dots ,x)$ satisfying $x^p=1$. Since $|X|\equiv 0(\mathrm{mod}p)$, the number of fixed points is divisible by $p$. The identity supplies one, so there is a nonidentity $x$ with $x^p=1$. Its order divides the prime $p$ and is not one, hence is $p$.

Cayley theorem states that every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. For $g\in G$, defineEach $L_g$ is a permutation, and $L_g{L}_h=L_{gh}$, so $g\mapsto L_g$ is a homomorphism. If $L_g$ is the identity permutation, then $g=L_g(1)=1$; hence the homomorphism is injective. For finite $G$ of order $n$, this embeds $G$ in $S_n$.

Five examples arewhere $D_{12}$ denotes the symmetry group of a hexagon and ${\mathrm{Dic}}_3$ is the $3$-dicyclic group.

The first two are abelian, while the last three are not. Of the first two, only $C_{12}$ is cyclic. Among the nonabelian groups, $A_4$ has no element of order $6$; both of the others do. The dihedral group $D_{12}$ has seven involutions, whereas ${\mathrm{Dic}}_3$ has the unique involution $a^3$ and also has elements outside $⟨a⟩$ of order four. These invariants prove that all five groups are pairwise non-isomorphic.

From $ba{b}^{-1}=a^{-1}$ we obtain $ba=a^{-1}b$. Moving every occurrence of $b$ to the right and replacing $b^2$ by $a^n$ shows that every group element has the formThe element $b$ cannot lie in $⟨a⟩$: otherwise it would commute with $a$, forcing $a=a^{-1}$, contrary to $a$ having order $2n>2$. Thus the two cosets $⟨a⟩$ and $⟨a⟩b$ are disjoint and each has $2n$ elements. Hence every $n$-dicyclic group has order $4n$.

Existence is explicit. Put $\zeta =e^{\pi i/n}$ and takeThen $a$ has order $2n$,The $2n$ diagonal matrices $a^k$ and the $2n$ off-diagonal matrices $a^{k}b$ are distinct, so they form a dicyclic group of order $4n$.

Let $\varphi :A_5\times A_5\to \{\pm 1\}$ be a homomorphism. Its restrictions to $A_5\times \{1\}$ and $\{1\}\times A_5$ cannot be nontrivial, because any nontrivial subgroup of $\{\pm 1\}$ is the whole group and part (iii) rules out such a surjection. Both restrictions are therefore trivial. Since$\varphi$ itself is trivial and in particular is not surjective.

If a homomorphism $A_5\to \{\pm 1\}$ were surjective, its kernel would be a normal subgroup of index two and hence order $30$. This contradicts the simplicity of $A_5$. Therefore no such surjective homomorphism exists.

The conjugacy classes in $A_5$ consist of the identity, the $15$ double transpositions, the $20$ three-cycles, and two classes of $12$ five-cycles. Thus their sizes areA normal subgroup is a union of conjugacy classes containing the identity, and its order must divide $60$ by Lagrange's theorem. Checking sums of a subset of $15,20,12,12$ shows that no numberdivides $60$. Hence the only possible normal-subgroup orders are $1$ and $60$. This proves the Simplicity of the alternating group A5.

For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and define the sign of a permutation bySince $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$,so this is a homomorphism. A transposition interchanges two columns of the identity matrix and therefore has determinant $-1$. If $\sigma$ is expressed as a product of $r$ transpositions, then $\mathrm{sgn}(\sigma )=(-1{)}^r$; because the determinant depends only on $\sigma$, the parity of $r$ is independent of the chosen expression.

For $F=(x,y,z)$,and all cross-partial terms in the curl vanish, so

The first identity in part (ii), with $F$ and $G$ interchanged, saysso the proposed identity is missing its final term. For a concrete counterexample, takeThen $\nabla \times G=0$, so the proposed left side is zero, whereasThus the identity is false.

Use the Levi-Civita symbol and the product rule. First,For the second identity, the contractiongivesIn vector notation this isThese are the divergence and curl of a cross product identities.

In the Cartesian basis, writing $F=(F_1,F_2,F_3)$,defines the divergence, while the curl isEquivalently,

Choose an angle $2\pi \alpha$ with $\alpha \notin \mathbb{Q}$. Rotation through that angle has infinite order, so the subgroup it generates is isomorphic to $\mathbb{Z}$.

The group also contains a copy of ${\mathbb{Z}}^2$. Choose real numbers $\alpha ,\beta$ such that $1,\alpha ,\beta$ are linearly independent over $\mathbb{Q}$, for example $\alpha =\sqrt{2}$ and $\beta =\sqrt{3}$. DefineThis is a homomorphism. If $\varphi (m,n)=I$, then $m\alpha +n\beta \in \mathbb{Z}$, and the stated rational independence forces $m=n=0$. Thus $\varphi$ is injective and its image is isomorphic to ${\mathbb{Z}}^2$.

The special orthogonal group isIfits first column is a unit vector, say $(\cos \theta ,\sin \theta {)}^T$. The second column is the unique unit vector perpendicular to it that gives positive determinant, namely $(-\sin \theta ,\cos \theta {)}^T$. Thusso $A$ is rotation through $\theta$ about the origin.

For $A\in SO(3)$, its real characteristic polynomial of odd degree has a real eigenvalue. Every eigenvalue of an orthogonal matrix has modulus one, so every real eigenvalue is $\pm 1$. Nonreal eigenvalues occur in conjugate pairs whose product is one, while $\det A=1$; if all eigenvalues are real, their product likewise forces at least one to be $+1$. Hence $A$ fixes a nonzero vector $u$. The plane $u^{\perp }$ is $A$-invariant, and the restriction to it is an orientation-preserving orthogonal map. By the $SO(2)$ result it is a planar rotation. Therefore $A$ is a rotation about the axis $\mathbb{R}u$.

Contracting two antisymmetric tensors gives the nonzero isotropic rank-four tensorA nonzero isotropic rank-five tensor is, for example,

More generally, one obtains the rank-five isotropic building blocks by choosing which two of the five indices belong to the Kronecker delta and putting the remaining three into the Levi-Civita symbol. There aresuch placements:Thus the unrestricted isotropic rank-five ansatz requires ten scalar component coefficients, one for each delta-epsilon placement.

Both sides are alternating in $i,j,k$ and in $l,m,n$. If either triple repeats an index, both sides vanish. Otherwise each triple is a permutation of $(1,2,3)$; the determinant of the corresponding permutation matrix is exactly the product of the two permutation signs. HenceThere are $3!=6$ nonzero components of ${ϵ}_{ijk}$, each equal to $\pm 1$, so