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Representations. A complex group representation is a group homomorphism $\rho :G\to \mathrm{GL}(W)$ to the invertible linear maps on a vector space. It is an irreducible representation when $W\ne 0$ and the only invariant subspaces are $0$ and $W$.

Group table. The supplied matrices satisfy $A=-I$, $C=-B$, and $A^2=B^2=C^2=I$. Their multiplication table isThe table proves closure and gives identity $I$ and an inverse for every element; associativity is inherited from matrix multiplication. The table is symmetric, so the group is abelian. It is the Klein four-group $V\cong C_2\times C_2$.

Dimensions and conjugacy. The sum of squares of irreducible representation dimensions is $|V|=4$. A two-dimensional irreducible would alone contribute four, leaving no room for the ever-present one-dimensional trivial representation. Hence the given two-dimensional representation is reducible. Since the group is abelian, every conjugacy class is a singleton. There are therefore four inequivalent complex irreducible representations, each of dimension one:Taking the trace of the given matrices gives their character of a representation:

Irreducible characters. In a one-dimensional representation, the values on $A$ and $B$ must each be $\pm 1$, since both elements square to $I$. Their values determine the value on $C=AB$. The four choices give the complete character table:Each row is a character because it respects the generating relations, and the rows have unit norm and pairwise zero character inner product. The multiplicities of these rows in $(2,-2,0,0)$ are respectively $0,0,1,1$, by character orthogonality. Thus

Diagonal form. The common invariant lines are spanned by $(1,1{)}^T$ and $(1,-1{)}^T$. Use their normalized vectors as columns ofThen simultaneous diagonalization givesReading the first and second diagonal entries across the four matrices recovers ${\chi }_3$ and ${\chi }_4$, respectively, verifying the decomposition explicitly.

Definitions and labels. A complex group representation on a vector space $V$ is a group homomorphism $\rho :G\to \mathrm{GL}(V)$, so $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. It is an irreducible representation if $V\ne 0$ and its only invariant subspaces are $0$ and $V$.

Read permutation matrices by their action on column basis vectors. The labels in the original PDF areIn particular, the original PDF gives the last row of $C$ as $(1,0,0)$; the provided TeX transcription's $(0,0,1)$ is incorrect and would not give an invertible permutation matrix.

Multiplication and conjugacy. Matrix products act on columns with the right factor first. For example $A^2=B$, $AD=C$, and $DA=E$. The complete multiplication table, with row factor multiplied by column factor, isConjugating a permutation relabels its cycles. Thus the conjugacy classes areof sizes one, two and three. These are respectively the identity, the three-cycles and the transpositions in the symmetric group on three letters.

Dimensions. The sum of squares of irreducible representation dimensions equals the group order, which here is six. A three-dimensional irreducible representation would already contribute nine, so the displayed representation is reducible. There are three inequivalent complex irreducible representations, one for each conjugacy class, including the trivial representation of dimension one. If their dimensions are $1,d_2,d_3$, then $d_2^2+d_3^2=5$, whose positive integer solution is $\{d_2,d_3\}=\{1,2\}$. HenceOne can also see reducibility directly: every permutation matrix fixes $(1,1,1{)}^T$.

Characters. The character of a representation is the trace of its representing matrix, constant on each conjugacy class. The given permutation representation has character equal to the number of fixed basis vectors:in class order $(I;A,B;C,D,E)$. A trivial representation has character $(1,1,1)$. For the second one-dimensional character, a transposition has value $\pm 1$ because its square is the identity. The two three-cycles are conjugate and inverse, so their common value $a$ obeys both $a^2=1$ and $a^3=1$, forcing $a=1$. Character orthogonality with the trivial character then requires $1+2+3b=0$, giving $b=-1$. This determines the sign representation without constructing its matrices.

Write the two-dimensional character as $(2,x,y)$. Character orthogonality with the first two rows givesso $x=-1$, $y=0$. Its squared norm is $(4+2)/6=1$, as required. The complete character table is thereforeThe last row belongs to the standard representation of the symmetric group.

To decompose the given representation, use the character inner productIts multiplicities areThusConcretely the invariant line is spanned by $(1,1,1{)}^T$ and its invariant complement is $\{(x,y,z):x+y+z=0\}$. Their characters are exactly the first and third rows, agreeing with the decomposition obtained entirely from character orthogonality.

The two one-dimensional representations are the trivial representation, ${\rho }_{+}(g)=1$, and the sign representation, ${\rho }_{-}(g)=\mathrm{sgn}(g)$. The first respects multiplication immediately. For the second, the sign of a permutation is the determinant of its permutation matrix, and multiplicativity of the determinant proves ${\rho }_{-}(gh)={\rho }_{-}(g){\rho }_{-}(h)$. They are inequivalent since a transposition has images $1$ and $-1$, and conjugation by a nonzero scalar cannot change a one-dimensional matrix.

Their group homomorphism kernels areBoth are subgroups, indeed normal subgroups, because they are kernels. In the second case $A_4$ is the alternating group of even permutations, with twelve elements.

The characters are their scalar values, since the trace of a one-dimensional matrix is its only entry. In the class order from part (b),The character inner product is the group average of ${\chi }_{+}\overline{{\chi }_{-}}$, soEach has squared norm $(1+6+3+8+6)/24=1$, in agreement with character orthogonality for distinct irreducible representations.

For a one-dimensional character, write $\chi (i)=ϵ$ and $\chi (j)=\eta$. Since $\chi (-1)=1$, the relations $i^2=j^2=-1$ require $ϵ,\eta \in \{\pm 1\}$. Also $ij=k$ gives $\chi (k)=ϵ\eta$. The four sign choices therefore supply all four linear characters.

For the remaining two-dimensional character ${\chi }_2$, its value at the identity is two. Column orthogonality between the identity and $-1$ gives $4+2{\chi }_2(-1)=0$, so ${\chi }_2(-1)=-2$. Between the identity and each of the other classes, the four linear-character values sum to zero, forcing the remaining character values to vanish. Thus the character table isHere column orthogonality is the standard finite-group character theorem ${\sum }_{\rho }{\chi }_{\rho }(e{)}^{\ast }{\chi }_{\rho }(g)=0$ for $g\ne e$. The rows also have unit norm and are mutually orthogonal in the character inner product with the displayed class-size weights. This completes the character table of the quaternion group.

Taking traces of the given matrices gives the character on the five classes,For a complex finite-group representation, the multiplicity of an irrep $\rho$ is its character inner product with ${\chi }_D$:Every linear character has value one at both $1$ and $-1$, so its multiplicity is $(4-4)/8=0$. For the two-dimensional character,ThereforeThe dimensions check: $2\times 2=4$.

The choice of scalar field matters here, even though the displayed matrices are real. Over $\mathbb{R}$, this particular four-dimensional representation is irreducible: for any nonzero real vector $v$, the vectors $v,D(i)v,D(j)v,D(k)v$ are mutually orthogonal with the same nonzero norm. This follows from the skew-symmetry and quaternion multiplication identities of the three matrices. A nonzero invariant real subspace contains all four vectors and is therefore the whole space. Its complexification is the two-copy decomposition above.

The rank is the number of orbits on ordered pairs under the diagonal group action. Its fixed-pair count at $g$ is $\chi (g{)}^2$, since the permutation character counts fixed points. Burnside's lemma identifies its average with the orbit count, while real-valuedness makes that average the character inner product of $\chi$ with itself. For a transitive action on at least two points, rank two is equivalent to transitivity on distinct ordered pairs.

Over the complex numbers, the signed Young permutation module isThe isomorphism is the tensor identity for induced representations. The partition index is unchanged by this twist; conjugation instead changes the labels of its irreducible constituents. For the conjugate partition ${\lambda }^{\prime }$, the character inner product of $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ is one: Mackey restriction formula reduces it to the unique zero-one matrix with row margins $\lambda$ and column margins ${\lambda }^{\prime }$.