Divide by the leading coefficient for $x\ne 0,1$:The pole of $P$ makes $x=0$ a singular point of a differential equation. However,are real analytic at zero. ThereforeThe ordinary-point power-series theorem does not apply there, but the Frobenius method does.

Set $c=\cos (2{\alpha }_1)$ and $s=\sin (2{\alpha }_1)$. Multiplying the matrices, or using the Pauli matrix multiplication law, givesCollecting the coefficients of $i{\tau }_j$ in $M^{\prime }$ yieldsThis is the right-handed active rotation by $2{\alpha }_1$ about the positive $x_1$ axis. Likewise, with $c_j=\cos (2{\alpha }_j)$ and $s_j=\sin (2{\alpha }_j)$,Thus the other axes also have rotation angle $2{\alpha }_j$. The factor of two is essential: replacing $A_j$ by $-A_j$ leaves the conjugation unchanged, illustrating the two-to-one relation between SU(2) and three-dimensional rotations.

Writing the matrix explicitly givesIts determinant isSince $\det A_1^{†}=(\det A_1{)}^{\ast }=1$,The conjugation therefore preserves the squared Euclidean length encoded in $M$.

For $A_1$, the determinant is ${\cos }^2{\alpha }_1-(i\sin {\alpha }_1{)}^2=1$. For $A_2$ it is ${\cos }^2{\alpha }_2+{\sin }^2{\alpha }_2=1$, and for $A_3$ it is $e^{i{\alpha }_3}{e}^{-i{\alpha }_3}=1$. Multiplicativity of the determinant givesTogether with unitarity, this places the matrices and their product in SU(2).

Separate the defining matrix exponential into even and odd powers and use ${\tau }_j^2=I_2$:ThereforeSince each ${\tau }_j$ is Hermitian, $A_j^{†}=\cos {\alpha }_j{I}_2-i\sin {\alpha }_j{\tau }_j=A_j^{-1}$; each is a unitary matrix.

Direct matrix multiplication givesFor example, the diagonal entries of ${\tau }_2^2$ are $(-i)i=i(-i)=1$, and its off-diagonal entries vanish. Associativity now reduces every positive integer power to its integer parity:These are the standard Pauli matrices, each a Hermitian matrix and an involution.

For a finite group over the complex numbers, the sum of the squared dimensions of its inequivalent irreducible representations equals the group order. There is always a one-dimensional trivial representation, and part (b) supplies an irreducible representation of dimension three. Thus the remaining squared dimensions sum toThey must be two further dimensions equal to one. HenceThe number of inequivalent irreducible representations equals the number of conjugacy classes, so $A_4$ has four classes. There are only three cycle types in part (a); consequently cycle type cannot classify conjugacy within $A_4$.

More explicitly, the identity is a singleton class, and conjugation by three-cycles permutes the three double transpositions transitively. For a three-cycle, the centralizer in $A_4$ consists of its three powers, so its conjugacy class has $12/3=4$ elements. The eight three-cycles therefore split into two classes of four:For example, $(123)$ and $(132)$ have the same cycle type but are not conjugate in $A_4$. Any permutation conjugating the first to its inverse must fix its unique fixed vertex and reverse the orientation of the remaining three labels, so it is odd. They are conjugate in $S_4$, illustrating exactly why restricting the ambient group changes the answer.

A rigid rotation of a regular tetrahedron permutes its four vertices, giving an injective group homomorphism from its rotation group to $S_4$: fixing every vertex forces the rotation to be the identity. Its seven nontrivial rotation axes are four lines through a vertex and the opposite face centre, and three lines through midpoints of opposite edges.

Each vertex-face axis admits rotations by $\pm 2\pi /3$, giving eight three-cycles on the vertices. Each opposite-edge axis admits a half-turn, giving three double transpositions. Along with the identity there are twelve rotations. All their induced permutations are even, so the injective image is a twelve-element subgroup of $A_4$, hence the whole group. ThereforeIts action on ${\mathbb{R}}^3$ supplies a three-dimensional representation. To prove irreducibility even after complexification, choose verticesThe group contains the half-turns $D_x=\mathrm{diag}(1,-1,-1)$ and their cyclic variants, as well as the coordinate cycle $(x,y,z)\mapsto (y,z,x)$. If a nonzero invariant subspace over the complex numbers contains a vector with nonzero $x$ component, applying $(I+D_x)/2$ shows it contains the $x$ axis. If another component is nonzero, use the corresponding projection instead. The coordinate cycle then supplies the other two axes. Every nonzero invariant subspace is therefore the full space, provingThis also gives an elementary proof without presupposing the character table.

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

The radial unit vector in the spherical coordinate system isThe coordinate tangent in the polar direction has length $r$, so its corresponding unit vector isDifferentiating ${\mathbf{e}}_r$ at fixed $\varphi$ therefore givesThe factor is not $r$: the corresponding scale factors of orthogonal coordinates occur in the derivative of the position vector, whereas the question differentiates its unit direction. The formula is understood away from the coordinate singularities where a complete spherical basis is not uniquely specified.

The kernel of a group homomorphism found in part (b) is the translation subgroupIt is a subgroup because $T_c{T}_d=T_{c+d}$, $T_0=I$ and $T_c^{-1}=T_{-c}$. It is nontrivial since $T_1\ne I$, and proper since, for example, $M(2,0)\notin N$.

Every kernel is a normal subgroup: if $n\in \ker f$, then $f(gn{g}^{-1})=f(g)ef(g{)}^{-1}=e$. Applying the same argument with $g^{-1}$ gives equality of the conjugate subgroup and the kernel. Here one can also check invariance under conjugation directly:Multiplication by nonzero $a$ permutes all $c\in \mathbb{C}$, so $M(a,b)NM(a,b{)}^{-1}=N$. Thus $N$ is a nontrivial proper normal subgroup, isomorphic to the additive group $(\mathbb{C},+)$.

Write $M(a,b)=\left(\begin{array}{cc}a& b\\ 0& 1\end{array}\right)$. Direct matrix multiplication givesSince $a,c\ne 0$, the product is again in $G$. Associativity is inherited from matrix multiplication, the identity element is $M(1,0)$, andlies in $G$. These verify all group axioms. This is the Affine group of the complex line, acting by $z\mapsto az+b$.

For the diagonal subset, $M(a,0)M(c,0)=M(ac,0)$, it contains $M(1,0)$, and $M(a,0{)}^{-1}=M(a^{-1},0)$. Thus $G^{\prime }$ is a subgroup. Finally,so $f:G\to G^{\prime }$ is a surjective group homomorphism.