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The quaternion group has conjugacy classes $\{1\}$, $\{-1\}$, $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$, with sizes $1,1,2,2,2$. Over the complex numbers, the number of classes and the sum of squares of irreducible representation dimensions force four one-dimensional representations and one two-dimensional representation. For $ϵ,\eta \in \{\pm 1\}$, the linear character rows on these classes are $(1,1,ϵ,\eta ,ϵ\eta )$. Orthogonality then gives the remaining row $(2,-2,0,0,0)$. A representation with character $(4,-4,0,0,0)$ contains two copies of the two-dimensional irrep and none of the linear irreps, by the character inner product. The choice of scalar field matters: the corresponding quaternionic four-dimensional real representation is irreducible over the reals.

For $C_n=⟨g:g^n=e⟩$, the complex irreducible representations are precisely the $n$ one-dimensional maps $g\mapsto e^{2\pi ij/n}$, with $0\le j<n$. Their characters have the displayed values, forming the discrete Fourier character table. Indeed a representing matrix $A$ satisfies $A^n=I$. Applying the fundamental theorem of algebra to its characteristic polynomial gives an eigenvalue and an eigenvector over the complex numbers. Its line is an invariant subspace under every power of $A$; irreducibility therefore forces the entire space to be that line. Its eigenvalue is an $n$th root of unity. Different roots give inequivalent one-dimensional representations, and there are exactly $n$ possibilities.

All conjugacy classes are singletons because the group is Abelian. The number of the displayed irreducibles equals the group order, and their squared dimensions sum to $n$, in agreement with the sum of squares of irreducible representation dimensions. Over the real numbers, conjugate nonreal characters instead combine into the planes described by the irreducible real representations of a finite cyclic group.

The matrix $r$ has distinct eigenvalues $\alpha ,{\alpha }^2$. Any one-dimensional invariant subspace for the defining representation would therefore have to be one of its two coordinate eigenspaces. But $s$ exchanges those lines, so neither is invariant under the whole group. Hence $R_2$ is an irreducible representation over the complex numbers.

There are three complex irreducible representations up to equivalence of group representations, because there are three conjugacy classes. The trivial representation and the sign representation already give two inequivalent one-dimensional representations. Together with $R_2$, their dimensions satisfy the sum of squares of irreducible representation dimensions:Thus there are no further irreducible representations. Alternatively the character of $R_2$ is $(2,-1,0)$, and its character inner product with itself is $(4+2)/6=1$, independently confirming irreducibility.

There are three complex irreducible representations, because there are three conjugacy classes. Two are the trivial representation $1$ and the sign representation $\epsilon$: both send $r$ to $1$, while they send $s$ to $1$ and $-1$, respectively. These exhaust the one-dimensional representations, since their scalar images must satisfy $r=r^{-1}$ and $r^3=1$, forcing $r=1$, and then $s^2=1$.

The defining two-dimensional matrix representation $D$ has characterIts character inner product with itself isso it is an irreducible representation. Directly, the two coordinate lines are the distinct eigenspaces of $r$, and $s$ exchanges them, leaving no common invariant line. The sum of squares of irreducible representation dimensions is $1^2+1^2+2^2=6$, proving completeness. In the conjugacy class order of solution, the character table isRepresentatives $r,s$ in this table stand for their full conjugacy classes. This lists all irreducible representations up to equivalence of group representations.

Write the two Pauli matrices asThey satisfy $A^2=B^2=I$, $BA=-AB$ and $C^2=-I$. Moving every $A$ to the left of every $B$ reduces each word to one of eight distinct matrices. ConsequentlyTaking $r=C$, $s=A$ gives $r^4=s^2=I$ and $srs=r^{-1}$, identifying the group with the dihedral group of a square.

The elements $\pm I$ are central. Conjugating $A$ by $B$ changes its sign, and similarly for $B$ and $C$. No conjugation mixes these pairs: modulo the central subgroup $\{\pm I\}$ the quotient group is the Klein four-group and hence Abelian. Thus the conjugacy classes are preciselyThere are five complex irreducible representations, one for each conjugacy class. In a one-dimensional representation, the images of $A,B$ are signs $s,t\in \{1,-1\}$. Since they commute, $-I$ maps to $1$, and all four choices of $s,t$ satisfy the relations. These are four inequivalent one-dimensional representations. The sum of squares of irreducible representation dimensions leaves $8-4=4$ for the fifth, so the dimensions are

The symmetric group $S_3$ has three conjugacy classes, determined by cycle type: the identity, three transpositions, and two three-cycles. Hence it has three complex irreducible representations. Two are the trivial representation $1$ and the sign representation $\epsilon$. Their dimensions are one, so the sum of squares of irreducible representation dimensions gives $1+1+d^2=6$, whence $d=2$ for the remaining representation $D$.

Write ${\chi }_D=(2,a,b)$ on these classes. Orthogonality to $1$ and $\epsilon$ gives $2+3a+2b=0$ and $2-3a+2b=0$. Thus $a=0,b=-1$, and the character table isThe character of a tensor product representation is the product of the factor characters. Thus ${\epsilon }^2=1$, $\epsilon \otimes D=D$, and ${\chi }_{D\otimes D}=(4,0,1)={\chi }_1+{\chi }_{\epsilon }+{\chi }_D$. These determine all the Clebsch-Gordan decomposition of group representations:

In a one-dimensional representation, the scalar images of $Q$ and $P$ commute. The relation $PQP=Q^{-1}$ gives $R(Q{)}^2=1$, while $R(Q{)}^5=1$, so $R(Q)=1$ and $R(P)=\pm 1$. These give exactly $R_0$ and $R_1$.

For $k=1,2$, define the two-dimensional representationsThe same invariant subspace argument as in solution proves that each is irreducible. Set $R_2=D_1$ and $R_3=D_2$. Their characters differ at $Q$, so they are inequivalent. The sum of squares of irreducible representation dimensions is already $1^2+1^2+2^2+2^2=10$, proving completeness.

Put $s=2\cos (2\pi /5)=(\sqrt{5}-1)/2$ and $t=2\cos (4\pi /5)=-(\sqrt{5}+1)/2$. Taking traces yields the character tableThe identities $s+t=-1$, $st=-1$, and $s^2+t^2=3$ also verify character orthogonality directly.

The trivial representation and sign of a permutation giveThey are one-dimensional representations, hence irreducible representations. Their characters, in the conjugacy class order of solution, areThese are the only one-dimensional representations. All transpositions are conjugate, so they have the same scalar image $t$ in any one-dimensional representation; since a transposition squares to the identity, $t=\pm 1$. Transpositions generate the symmetric group, fixing the whole representation once $t$ is chosen.

By the sum of squares of irreducible representation dimensions, the three remaining dimensions satisfy $d_2^2+d_3^2+d_4^2=24-1-1=22$, with each $d_i\ge 2$. No $d_i$ can be at least $4$, since $16+4+4>22$. Thus the only possible dimensions are

There is one complex irreducible representation for each conjugacy class, hence three. The trivial representation $1$ and sign representation $\epsilon$ are distinct and one-dimensional. The sum of squares of irreducible representation dimensions leavesCall the remaining irreducible representation $2$. The permutation representation on ${\mathbb{C}}^3$ has character equal to the number of fixed objects, namely $(3,1,0)$ on the ordered conjugacy classes $(e,\text{transpositions},\text{three-cycles})$. Removing its constant-vector trivial representation gives $(2,0,-1)$, whose character inner product with itself is one. This constructs $2$ and gives the character tableTo obtain every Clebsch-Gordan decomposition of group representations, multiply the characters pointwise. In particular, ${\chi }_{\epsilon }^2={\chi }_1$, ${\chi }_{\epsilon }{\chi }_2={\chi }_2$ andThus all ordered pairs of irreducible representations are covered by

Put $r=g_1$ and $s=g_2$. Direct matrix multiplication givesMoving every $s$ to the right shows that every word in these generators of a group has the form $r^a{s}^b$, with $a\in \{0,1,2\}$ and $b\in \{0,1\}$. The three $r^a$ are distinct diagonal matrices, and the three $r^{a}s$ are distinct off-diagonal matrices. HenceSince $srs=r^2\ne r$, this is a non-abelian group. Its group presentation identifies it with the dihedral group of an equilateral triangle, equivalently the symmetric group $S_3$.

The conjugacy classes areIndeed $s$ interchanges $r$ and $r^2$ by conjugation, while $r^{a}s{r}^{-a}=r^{2a}s$ runs through all three reflections. No reflection is conjugate to a rotation, since conjugation preserves order of a group element.

There are three inequivalent complex irreducible representations, one for each conjugacy class. The sum of squares of irreducible representation dimensions gives $d_1^2+d_2^2+d_3^2=6$. One representation is trivial, and the only possible positive integer dimensions are

Write $1$ for the trivial representation, $\epsilon$ for the sign representation, and $2$ for the given two-dimensional group representation. Their matrix representations on the generators of a group arewith $\alpha =e^{2\pi i/3}$. The two eigenvalues of $r$ in $2$ are distinct, and $s$ exchanges its two eigenlines; no line is invariant under both. Thus $2$ is an irreducible representation. The dimensions already exhaust the sum of squares of irreducible representation dimensions, so this is the complete list.

Since $\alpha +{\alpha }^{-1}=-1$, the character table isMultiplying characters and using character orthogonality gives every tensor product of group representations:For example ${\chi }_2^2=(4,1,0)={\chi }_1+{\chi }_{\epsilon }+{\chi }_2$. Explicit Clebsch-Gordan coefficients can be read from two doublets $x=(x_1,x_2{)}^T$, $y=(y_1,y_2{)}^T$:The first two combinations are invariant under $r$ and respectively even and odd under $s$; the last transforms by the displayed doublet matrices.

Put $\zeta =e^{\pi i/n}$. The group representations from the preceding construction with $ϵ={\zeta }^j$, $1\le j\le n-1$, will be denoted ${\rho }_j$. Their diagonal matrices for $a$ have distinct eigenvalues. The only possible invariant lines would be these two eigenspaces, and the matrix for $b$ exchanges them. Thus each ${\rho }_j$ is irreducible. Equivalently their $a,b$ matrices do not commute, so part (a) applies. The choices $j$ and $2n-j$ give equivalent group representations by exchanging basis vectors. The displayed range contains exactly one of each pair, and distinct members have different traces at $a$ because $\cos (\pi j/n)$ is strictly decreasing there.

For a one-dimensional character of a representation, write $a\mapsto \alpha$, $b\mapsto \beta$. The conjugation relation forces ${\alpha }^2=1$, and ${\beta }^2={\alpha }^n$. Hence there are exactly four choices:For even $n$, $\beta =\pm 1$ for either $\alpha$; for odd $n$, $\alpha =1$ has $\beta =\pm 1$ and $\alpha =-1$ has $\beta =\pm i$. Finally,The sum of squares of irreducible representation dimensions proves that these four one-dimensional and $n-1$ two-dimensional group representations exhaust the irreducibles, including $n=1$ where the latter list is empty.

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.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$. It is a reducible representation if there is a nonzero proper invariant subspace of $V$ preserved by every $D(g)$, and an irreducible representation otherwise. With $S$ invertible,so changing basis gives another representation. Its character of a representation is $\chi (g)=\mathrm{tr}D(g)$, and cyclicity of the trace gives $\tilde{\chi }(g)=\mathrm{tr}(SD(g)S^{-1})=\chi (g)$.

A conjugacy class consists of $hg{h}^{-1}$ as $h$ varies. Since $(hg{h}^{-1}{)}^k=h{g}^k{h}^{-1}$, a power equals the identity exactly when $g^k$ does. Thus conjugation preserves element order, including infinite order.

For the square, let $r$ be rotation by $\pi /2$ and $s$ reflection in a horizontal symmetry axis. The dihedral group has the eight elementsThe conjugacy classes of a finite dihedral group specialize toThe two central rotations are singletons. Reflection conjugation reverses rotations, while conjugation by $r$ shifts a reflection exponent by two, giving the two reflection classes.

For finite groups over $\mathbb{C}$, character orthogonality and completeness make the irreducible characters a basis of the class functions. The latter space has one independent value per conjugacy class. Hence the square group has five inequivalent irreducible complex representations. The sum of squares of irreducible representation dimensions is eight, forcing four one-dimensional representations and one two-dimensional representation. The four scalar choices $r\mapsto \pm 1$, $s\mapsto \pm 1$ supply the former; the ordinary action on the plane supplies the latter.

A group representation on a vector space $W$ is a group homomorphism $\rho :G\to GL(W)$: $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. An invariant subspace $U$ obeys $\rho (g)U\subseteq U$ for every $g$. A nonzero representation is irreducible if its only invariant subspaces are $0$ and $W$. It is a faithful representation if it is injective, equivalently if its group kernel is $\{e\}$. Its character is ${\chi }_{\rho }(g)=\mathrm{tr}\rho (g)$. For inequivalent complex irreducible representations of a finite group, character orthogonality is

Use right-to-left composition of permutations and assign the matrix labels consistently byThe resulting Cayley table, with row multiplying column, isConjugation relabels the entries of a cycle, preserving its length. Thus the conjugacy classes are

Define the permutation representation by $P_{\sigma }{e}_j=e_{\sigma (j)}$, or $(P_{\sigma }{)}_{ij}={\delta }_{i,\sigma (j)}$. Applying it to a basis vector proves $P_{\sigma }{P}_{\tau }=P_{\sigma \tau }$. The six assignments above produce exactly the listed matrices. They are distinct, and $P_{\sigma }=I$ forces $\sigma (j)=j$ for every $j$, so the representation is faithful.

Every permutation fixes $(1,1,1)$, hence fixes every vector $(c,c,c)$. Therefore $V_0$ is a one-dimensional invariant subspace carrying the trivial representation. Its orthogonal complement issince the inner product with $(c,c,c)$ is $c(v_1+v_2+v_3)$. Permuting coordinates preserves their sum, so this plane is invariant too. Every real vector has the unique decompositionThus the permutation representation is the direct sum of group representations on the line and the plane.

To find the eigenvectors of $A$, note that $A(v_1,v_2,v_3)=(v_3,v_1,v_2)$. With $\zeta =e^{2\pi i/3}$, its three eigenvalues and eigenspaces areThe only real eigenvalue is one, whose eigenline is $V_0$, so $A$ has no nonzero real eigenvector in $V_0^{\perp }$. A proper nonzero real invariant subspace of that two-dimensional plane would be a line, giving such an eigenvector. Hence the plane representation is irreducible over $\mathbb{R}$.

For the later complex dimension-count theorem one should also establish irreducibility over $\mathbb{C}$; real irreducibility alone would not imply it in general. In the complexified plane, the only candidate $A$-invariant lines are its $\zeta$ and ${\zeta }^2$ eigenlines. But $C$ exchanges them:Neither is invariant under the whole group, proving complex irreducibility as well. The required decomposition is thereforewhere $W$ is the standard representation of the symmetric group for three letters. Its character is obtained by subtracting one from the number of fixed coordinates: on the three classes it is $(2,-1,0)$.

Finally, the sign representation gives a second one-dimensional irreducible, with value $+1$ on $I,A,B$ and $-1$ on the transpositions. These are exactly the two one-dimensional possibilities: if $r=A,s=C$, a one-dimensional representation sends them to commuting nonzero scalars $v,u$. The relations $r^3=s^2=e$ and $srs=r^{-1}$ imply $v^3=u^2=1$ and $v=v^{-1}$. Hence $v=1$ and $u=\pm 1$, the trivial and sign choices. Together with the two-dimensional $W$, these giveThe sum of squares of irreducible representation dimensions leaves room for no other irreducible. Thus there are exactly two one-dimensional and one two-dimensional complex irreducible representations.

A group homomorphism $D:G_1\to G_2$ satisfies $D(gh)=D(g)D(h)$ for all $g,h\in G_1$. Its group kernel is

For the matrices, put $X=R(x)$, $Y=R(y)$ and $z=e^{2\pi i/3}$. Since $z^3=1$, $z\ne 1$ and $1+z+z^2=0$, direct matrix multiplication givesThese verify the listed matrices and the defining relations, with $R(e)=I_2$. To check every product explicitly through a single calculation, write elements in the unique form $x^{ϵ}{y}^j$, with $ϵ=0,1$ and $j=0,1,2$. The relations giveThe displayed matrices obey $Y^{j}X=X{Y}^{-j}$, so their product has exactly the same form:All exponents are reduced modulo two or three respectively. Thus this verifies all 36 products, and $R$ is a two-dimensional complex group representation. It is given to be an irreducible representation.

For a one-dimensional complex group representation $U$, let $u=U(x)$ and $v=U(y)$, which are nonzero complex numbers. The relations requireSince scalar multiplication commutes and $u\ne 0$, the last relation gives $v=v^2$, hence $v=1$. Then $u=\pm 1$. The choice $u=1$ is the trivial representation; the unique nontrivial choice is the sign representation:Its group kernel isA faithful representation is injective and has trivial kernel. Here the kernel has three elements, so $U$ is not faithful.

Both one-dimensional representations are irreducible representations, and they are inequivalent to each other and to the given two-dimensional one. The sum of squares of irreducible representation dimensions givesThese exhaust all complex irreducible representations. Their number equals the number of conjugacy classes, so there are three conjugacy classes. Explicitly they areThe last two sets consist of the 3-cycles and transpositions respectively.

The conjugacy class of $g\in G$ is $\{hg{h}^{-1}:h\in G\}$. An irreducible representation acts on a nonzero vector space with no nonzero proper subspace invariant under every group element.

In the quaternion group, $1$ and $-1$ commute with all elements, so each forms its own conjugacy class. The element $i$ commutes with $\pm 1,\pm i$, while conjugation by $j$ or $k$ changes it to $-i$. Thus its class is $\{i,-i\}$. The same argument for $j,k$ gives all five classes:Their sizes sum to eight. Equivalently each noncentral element has a four-element centralizer and hence a two-element class.

Use the following finite-group representation theorems over $\mathbb{C}$: the number of inequivalent irreducible representations equals the number of conjugacy classes, and their dimensions obey the sum of squares of irreducible representation dimensions, $\sum d_j^2=|G|$. Thus there are five irreducibles, with ${\sum }_{j=1}^5{d}_j^2=8$. Starting with five dimensions equal to one uses five of those eight units; increasing one dimension from one to two uses exactly the remaining three, while any larger increase would exceed eight. ThereforeThese are the dimensions underlying the character table of the quaternion group. The scalar field matters: the quaternionic irreducible becomes four-dimensional over the reals. The class-count and dimension-sum statements used here have their standard complex meaning.

Write the finite cyclic group as $C_n=⟨g:g^n=e⟩$, the source's $Z_n$. It is Abelian, so $hg{h}^{-1}=g$ and every conjugacy class is a singleton: there are $n$ such classes.

Over the complex numbers, all its irreducible representations are one-dimensional. One direct proof is that the matrix representing $g$ has an eigenvector over $\mathbb{C}$; its line is preserved by every power of that matrix and hence by the whole group. Irreducibility forces the line to be the whole representation space. Its eigenvalue must be an $n$th root of unity.

The complex irreducible representations of a finite cyclic group are thereforeTheir generator values are distinct, so no two are equivalent. There are $n$ of them, and $n$ squared dimensions equal one, exhausting the sum of squares of irreducible representation dimensions: ${\sum }_j{d}_j^2=n=|G|$. Thus the list is complete.

For $n=3$, let $\omega =e^{2\pi i/3}$, so ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$. The character table isThe complex-field convention is essential. The preceding part used real matrices, but that does not make the one-dimensional classification valid over $\mathbb{R}$: conjugate nonreal characters combine into the two-dimensional irreducible real representations of a finite cyclic group. The character table and dimension-sum result used here have their usual complex meaning.

For a finite group over $\mathbb{C}$, the number of inequivalent irreducible representations is its number of conjugacy classes, and the sum of squares of irreducible representation dimensions equals the group order. There are therefore five irreducibles for $S_4$, with squared dimensions summing to $24$.

First establish that part (c) found every one-dimensional representation. In such a representation, conjugate elements have the same scalar image. All transpositions are conjugate and square to the identity, so they all have one common image $s$ with $s^2=1$. Transpositions generate $S_4$, hence $s=1$ gives exactly the trivial representation, while $s=-1$ gives exactly the sign representation. There are precisely two degree-one irreducibles.

The other three dimensions $d_1,d_2,d_3$ are integers at least two and satisfyNo one of them can be at least four, since $4^2+2^2+2^2=24>22$. They are therefore all two or three. Starting from three twos gives sum of squares $12$; each replacement of a two by a three adds $5$. To reach $22$ requires exactly two replacements. Thus the complete list isThe count of conjugacy classes guarantees that these are all the inequivalent complex irreducible representations, not just dimensions of a proposed incomplete list.