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Lagrange's theorem says that, for a finite group $G$ and subgroup $H$,Here $[G:H]$ is the number of left cosets. Every $g\in G$ lies in $gH$. If $gH$ and $g^{\prime }H$ meet, write $gh=g^{\prime }{h}^{\prime }$; then $g^{\prime }=gh(h^{\prime }{)}^{-1}$, so $g^{\prime }H=gH$. Hence the left cosets partition $G$. The bijection $h\mapsto gh$ makes each coset have $|H|$ elements. Adding their cardinalities proves Lagrange's theorem.

False. Any supposed kernel of a group homomorphism of order two would have a nonidentity central element by the introductory proof. Every element of order two in $S_3$ is a transposition, and none is central: conjugating $(ij)$ by a suitable three-cycle gives a different transposition. Thus the center of a group for $S_3$ is trivial, and such a kernel is impossible. This argument does not assume that an arbitrary order-two subgroup is normal.

True. Use the odd-dimensional orthogonal determinant splitting:Scalar matrices commute with every matrix, so $\Phi (AB)=\Phi (A)\Phi (B)$. Its output is orthogonal andEvery $B\in SO(3)$ satisfies $\Phi (B)=B$, proving surjectivity. The equation $\Phi (A)=I$ forces $A=(\det A)I$, whenceIn odd dimension $-I$ has determinant $-1$, which is the sign needed in this construction.

True. The sign of a permutation is a group homomorphism $\mathrm{sgn}:S_3\to \{1,-1\}$, as follows, for example, from the multiplicative determinant of permutation matrices. Its kernel of a group homomorphism is the alternating groupA transposition maps to $-1$, so this homomorphism is onto.

True. The determinant gives a group homomorphismFor an orthogonal matrix, $A^{T}A=I$ implies $(\det A{)}^2=1$, and a reflection has determinant $-1$, so the target has order two and the map is onto. Its kernel of a group homomorphism is exactlyThis is the defining special orthogonal group inside the orthogonal group.

Applying a permutation to the elements of an $r$-element subset preserves its cardinality. The identity fixes each subset andso this is the symmetric group action on subsets. For two $r$-element subsets $U,W$, choose a bijection $U\to W$ and a bijection between their complements; together they define a permutation of the full $n$-element set. Hence the action is transitive.

A member of the stabiliser subgroup must permute $U$ internally and its complement internally, with independent choices. ThusThis also agrees with the orbit-stabilizer theorem; $0!=1$ handles $r=n$.

The identity permutation fixes every multivariate polynomial. With permutation composition $(\sigma \tau )(i)=\sigma (\tau (i))$, substitution givesThus this is a left variable-permutation action on polynomials. The substitution acts on variable names; using the displayed definition avoids reversing the composition order.

For the given multivariate polynomial, the group orbit corresponds to the three partitions of four indices into two unordered pairs:These three polynomials are distinct, and every pairing occurs under a permutation. The orbit-stabilizer theorem givesThe eight elements of the stabiliser subgroup independently interchange the members of each pair, and may also interchange the two pairs.

This plane curve is an Archimedean spiral. Put ${\mathbf{e}}_r=(\cos u,\sin u,0)$ and ${\mathbf{e}}_{\theta }=(-\sin u,\cos u,0)$, so ${\mathbf{e}}_r^{\prime }={\mathbf{e}}_{\theta }$ and ${\mathbf{e}}_{\theta }^{\prime }=-{\mathbf{e}}_r$. ThenTherefore the curvature of a plane curve isThe parametrization is regular even at $u=0$, where the curvature of a plane curve is $2$.

Differentiate the unit tangent vector with $d/ds=v^{-1}d/du$ and $v^{\prime }=({\mathbf{x}}^{\prime }\cdot {\mathbf{x}}^{\prime \prime })/v$. This givesThe curvature of a space curve is $\kappa =|d\mathbf{t}/ds|$. Squaring the displayed expression and simplifying givesThe last equality is the cross product norm identity. Equivalently, the hinted vector triple product rewrites $d\mathbf{t}/ds=-{\mathbf{x}}^{\prime }\times ({\mathbf{x}}^{\prime }\times {\mathbf{x}}^{\prime \prime })/v^4$.

For a regular curve parametrized by $u$, put $v(u)=|{\mathbf{x}}^{\prime }(u)|>0$ and orient arc length in the direction of increasing $u$. Since $ds=|d\mathbf{x}|=vdu$, the unit tangent vector isRegularity is the condition that permits division by $v$.

Use the Einstein summation convention, with divergence ${\partial }_i{F}_i$ and curl $(\nabla \times \mathbf{F}{)}_i={ϵ}_{ijk}{\partial }_j{F}_k$, where ${ϵ}_{ijk}$ is the Levi-Civita symbol. For the first vector field,The derivative matrix is symmetric, so its contraction with the Levi-Civita symbol vanishes. For the second vector field,For the third vector field, $H_i={ϵ}_{ikl}{a}_k{x}_l/r$. Its divergence is zero because the Levi-Civita symbol contracts with symmetric products. For its curl, either contract two Levi-Civita symbols or use the vector triple product:ThusAll formulas hold on $r>0$, the domain where the radial unit vector is defined.

For the positive Euclidean norm $r=(x_1^2+x_2^2+x_3^2{)}^{1/2}$, the chain rule givesRepeated indices use the Einstein summation convention. The origin is excluded because the Euclidean norm is not differentiable there. Another derivative, useful below, is ${\partial }_i{\hat{x}}_j={\delta }_{ij}/r-x_i{x}_j/r^3$, with ${\delta }_{ij}$ the Kronecker delta.

Let $N$ be a normal subgroup and let $\pi :G\to G/N$ be the quotient group homomorphism. By the first part, $G=⟨a,b⟩$ for involutions $a,b$, so the quotient group is generated by $\pi (a),\pi (b)$, whose squares are the identity. If both images are nonidentity, they are involutions and the second part applies. If only one image is nonidentity, it generates $C_2$; choose it twice. Both images cannot be trivial because the quotient group is nontrivial. Hence every nontrivial quotient is dihedral in the stated extended convention. This is the nontrivial-quotient closure of two-involution groups.

Let $a,b$ be the generating involutions. If $a\ne b$, set $x=ab$ and $y=b$. Then $x\ne e$, so its order of a group element is at least two or infinite, andTherefore $x,y$ generate the group and satisfy the required dihedral inversion relation. If $a=b$, the group is $C_2$ and $x=y=a$ gives its permitted degenerate description. This proves the two-involution characterization of a dihedral group without excluding either $C_2$ or the infinite dihedral group.

The two-involution characterization of a dihedral group uses the extended convention that includes $C_2$ and the infinite dihedral group; the two generators need not be distinct. Put $a=xy$ and $b=y$. The inversion relation givesIf $a\ne e$, both $a,b$ are involutions and generate the original group. If $a=e$, then $x=y$ and the group is $C_2$; choose the same nonidentity element twice. Thus two elements of order two generate the group, including the degenerate case allowed here. The distinction between square equal to the identity and order of a group element exactly two is essential.