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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.

Extend $\sigma \in S_n$ to a permutation $\hat{\sigma }$ of $n+2$ letters that fixes the last two letters. Let $t=(n+1n+2)$, and let $ϵ(\sigma )$ be $0$ or $1$ according as $\sigma$ is even or odd. DefineThe two factors have disjoint supports and commute. Their signs cancel, so $\iota (\sigma )\in A_{n+2}$. Additivity of parity of a permutation modulo two givesThus $\iota$ is a group homomorphism. Its restriction to the first $n$ letters recovers $\sigma$, making it injective. Its image is therefore a subgroup isomorphic to $S_n$:This group embedding works for every positive $n$, including the trivial case $n=1$.

Here isotropic tensors are invariant under proper rotation matrices, $R^{T}R=I$ and $\det R=1$. The isotropic second-rank tensor and isotropic third-rank tensor forms in three dimensions arewith scalar coefficients $A,B$. One can see completeness directly. A half-turn about each coordinate axis kills every off-diagonal component of a rank-two tensor, and quarter-turns make its three diagonal components equal. For rank three, the half-turns require that each coordinate label occur an odd number of times in a nonzero component. Thus only permutations of $(1,2,3)$ survive. Quarter-turns interchange two labels while reversing one sign, forcing the surviving components to alternate with the parity of a permutation. This gives a multiple of the Levi-Civita symbol.

The tensor transformation law verifies invariance of these forms:The first equality uses orthogonality and the second uses the determinant formula. If reflections were also required while using the ordinary tensor transformation law, the rank-three coefficient would have to vanish. The proper-rotation convention is the one under which $B{ϵ}_{ijk}$ is an isotropic tensor.

The ball is unchanged by any rotation matrix, and rotating coordinates has Jacobian determinant one. Thus changing variables $\mathbf{y}=R\mathbf{x}$ givesThese moments are therefore isotropic tensor integrals. For the Cartesian moments of a three-dimensional ball, the second moment is $\alpha {\delta }_{ij}$. Taking its tensor contraction givesThe third moment vanishes: $\mathbf{x}\mapsto -\mathbf{x}$ preserves the ball and changes the sign of its integrand. Equivalently, its symmetry under index interchange excludes a nonzero multiple of the antisymmetric Levi-Civita symbol.

The fourth moment is symmetric under every permutation of its four indices. The general rank-four isotropic tensor form consequently has equal coefficients, yieldingFor example, comparing $T_{1122},T_{1212},T_{1221}$ shows that the three coefficients agree. Contracting $i=j$ and $k=l$ givesFinally the vector triple product identity gives $\mathbf{x}\times (\mathbf{\Omega }\times \mathbf{x})=r^2\mathbf{\Omega }-(\mathbf{\Omega }\cdot \mathbf{x})\mathbf{x}$. The first term integrates to $3\alpha \mathbf{\Omega }$ and the second to $\alpha \mathbf{\Omega }$, using the second moment. Hence

A group homomorphism $h:G\to F$ preserves multiplication: $h(g_1{g}_2)=h(g_1)h(g_2)$. It sends the identity element to the identity and inverse elements to inverse elements. Its group kernel is $K=\{g\in G:h(g)=e_F\}$. This contains $e_G$ and is closed under multiplication and inverse elements, so it is a subgroup. For $k\in K$ and $g\in G$,Thus $gK{g}^{-1}\subseteq K$; applying the same argument with $g^{-1}$ gives equality. Hence the group kernel is a normal subgroup.

Let $S_2=\{e,\tau \}$ with $\tau =(12)$. The parity of a permutation defines the desired map from $S_3$:The sign of a permutation is multiplicative, so parity adds modulo two under composition, exactly matching ${\tau }^2=e$. Thus this is a group homomorphism; it is surjective because a transposition maps to $\tau$. Its group kernel isThis is the alternating group on three objects, a normal subgroup of index two in $S_3$.

The sign homomorphism $\mathrm{sgn}:S_n\to \{+1,-1\}$ records parity of a permutation. The kernel of a group homomorphism in this case is the alternating group $A_n$, consisting of even permutations. For $n\ge 2$ it has index two in the symmetric group and order $n!/2$.

The cycle type is the multiset of lengths of the disjoint cycles, including fixed points if desired. For $A_4$ the complete count isA single transposition and a four-cycle are odd and are excluded. Thus