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The profinite group $G$ is compact, and the preceding part shows thatis its image. By the continuous image of a compact space theorem, the conjugacy class is compact. Since a profinite group is Hausdorff, every compact subset is closed, so ${\mathrm{Cl}}_G(h)$ is closed.

Irreducible degrees divide the order of a finite $2$-group and satisfyThe number of linear characters is $|G/G^{\prime }|$. For a nontrivial finite $2$-group this is at least $4$ unless the group is abelian, and every nonlinear degree is at least $2$. A degree $4$ constituent is therefore impossible in the nonabelian case. If there are $\ell$ linear and $m$ degree-two characters, thenThus the only degree collections arewith respectively $r=16,10,7$.

For $1^{16}$ take the cyclic group $G=C_{16}=⟨g⟩$. Every conjugacy class is a singleton, and its character table iswhere ${\zeta }_{16}=e^{2\pi i/16}$.

For $1^8,2^2$, take $G=D_8\times C_2$, where
$D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ and
$C_2=⟨z⟩$. On the five classesthe four linear characters of $D_8$, indexed by
$\epsilon ,\delta \in \{\pm 1\}$, and its degree-two character areThe ten classes of $G$ are $(C,1)$ and $(C,z)$. Its full table consists ofwhere $\chi$ runs over the above five rows and
$\eta \in \{1,-1\}$ runs over the two characters of $C_2$. This explicitly gives eight linear and two degree-two rows.

For $1^4,2^3$, take the dihedral groupOrder its seven classes asThe four linear rows areThe three degree-two rows, for $k=1,2,3$, areThe row norms and mutual inner products, weighted by the displayed class sizes, verify irreducibility and completeness.

Conjugation by a permutation merely permutes the transpositions. Their sum $C$ is therefore a conjugacy class sum and belongs to the center of an associative algebra $\mathbb{C}{S}_n$. Since the complex Specht module is an irreducible representation, Schur lemma says that $C$ acts on it as a scalar, say $c(\lambda )$.

Let $P_{\bullet }\to \mathbb{Z}$ be a projective resolution of the trivial $\mathbb{Z}G$-module. The projective-resolution definition of group cohomology isThis is independent, up to a natural isomorphism, of the chosen projective resolution.

The degreewise natural isomorphismscommute with the coboundary maps. Taking cohomology proves that group cohomology commutes with finite direct sums:

Now restrict $P_{\bullet }$ from $G$ to a subgroup $K$. The group ring $\mathbb{Z}G$ is free as a $\mathbb{Z}K$-module, so restriction carries free modules to free modules and projective modules to projective modules. Thus the restricted complex is a projective resolution of the trivial $\mathbb{Z}K$-module. For the coinduced module $Y={\mathrm{Hom}}_{\mathbb{Z}K}(\mathbb{Z}G,X)$, the Hom functor adjunction for a coinduced module gives an isomorphism of cochain complexesExplicitly, a map $F$ is sent to $p\mapsto F(p)(1)$; the inverse sends a $\mathbb{Z}K$-linear map $a$ to $p\mapsto (r\mapsto a(rp))$. Taking cohomology proves Shapiro's lemma:

For the conjugation module of a group ring $N=\mathbb{Z}{G}_{\mathrm{conj}}$, the basis $G$ is the disjoint union of its conjugacy classes. Hence $N$ is the direct sum of the integral permutation modules on those classes. The class of a representative $g_i$ is the transitive $G$-set $G/C_G(g_i)$, where $C_G(g_i)$ is its centralizer. Since $G$ is finite, this permutation module is both induced and coinduced from the trivial $C_G(g_i)$-module $\mathbb{Z}$. Applying group cohomology commutes with finite direct sums and Shapiro's lemma yields the group cohomology of a conjugation module:

The symmetric group $S_3$ has three conjugacy classes, represented by the identity, a transposition, and a three-cycle. Their centralizers are respectivelyFor any finite group $L$ acting trivially on $\mathbb{Z}$,because a group homomorphism sends an element of finite order to an element of finite order, while the additive group of the integers contains no nonzero torsion elements. ThereforeThe periodic resolution of a finite cyclic group alternates the maps $t-1$ and $1+t+\cdots +t^{m-1}$. After applying ${\mathrm{Hom}}_{\mathbb{Z}{C}_m}(-,\mathbb{Z})$ with the trivial action, these become alternately zero and multiplication by $m$, provingCombining this calculation with the supplied $H^2(S_3,\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$ gives

If $g\in Z(G)$, it commutes with each $x\in G$, so $x^{-1}gx=g$. Its conjugacy class is therefore $C(g)=\{g\}$.

Conversely, if $C(g)=\{g\}$, then $x^{-1}gx=g$ for every $x$. Multiplication by $x$ on the left gives $gx=xg$ for every $x$, which is exactly membership in the center of a group. Thus

Suppose $\sigma \ne e$. Choose a symbol $i$ moved by $\sigma$ and write $j=\sigma (i)\ne i$. Since $n\ge 3$, choose a third symbol $k$ distinct from $i,j$. Let $\tau =(jk)$, a transposition that fixes $i$. ThenThus ${\tau }^{-1}\sigma \tau \ne \sigma$, proving that the conjugacy class of any nonidentity permutation has more than one element. By part (a)(iii), no such permutation lies in the center of a group. The identity always does, soThe restriction on $n$ is necessary: $S_2$ is an abelian group and equals its own center.

The orbit of $x$ under conjugation is its conjugacy classIt contains $x$ by using $g=e$, so the conjugacy classes cover $G$. If $z=gx{g}^{-1}$, then $x=g^{-1}zg$, and the group action identity shows that conjugating either element further gives the same orbit; hence $C(z)=C(x)$. If $C(x)$ and $C(y)$ intersect at $z$, this gives $C(x)=C(z)=C(y)$. Thus distinct conjugacy classes are disjoint and the conjugacy classes form a partition of a set $G$.

For a nonabelian group, iv and iii imply $|G|=|G/A||A|\ge 4|A|$. Substituting $|A|\le |G|/4$ into i yieldsThe constant is sharp: the quaternion group has eight elements and the five conjugacy classes $\{1\}$, $\{-1\}$, $\{i,-i\}$, $\{j,-j\}$, $\{k,-k\}$. To verify this directly, $\pm 1$ are central elements, conjugation by $j$ sends $i$ to $-i$, and similarly for $j,k$; its relations $i^2=j^2=k^2=-1$ and $ij=k=-ji$, $jk=i=-kj$, $ki=j=-ik$ show that conjugating $i,j,k$ by any of $\pm 1,\pm i,\pm j,\pm k$ changes only its sign. Hence $m(G)=5=(5/8)|G|$.

A conjugacy class $C(x)$ is a singleton exactly when $gx{g}^{-1}=x$ for every $g\in G$, equivalently when $gx=xg$ for every $g$. Thus the singleton conjugacy classes correspond exactly to elements of the center of a group $A$, and there are $|A|$ of them.

Every other conjugacy class has at least two elements. Since the conjugacy classes form a partition of a set $G$, if $r$ is the number of these remaining classes then $2r\le |G|-|A|$. ConsequentlyThis conjugacy class count bound from the center needs no classification of the finite group.

Assume first $n\ge 2$. All transpositions form a single conjugacy class $T_1$. A group automorphism preserves order of a group element, so the image of a transposition is a nonidentity involution. By the order formula, its disjoint cycle decomposition contains only one-cycles and two-cycles, with at least one two-cycle. Thus it lies in some $T_k$, where $1\le k\le \lfloor n/2\rfloor$.

Moreover an automorphism maps the entire conjugacy class onto the conjugacy class of the image: for a fixed transposition $t$,and $\varphi (g)$ runs over all of $S_n$ by surjectivity. Part (a)(i) identifies that latter conjugacy class as $T_k$. Hence $\boxed{\varphi (T_1)=T_k}$, with equality rather than just inclusion. If $n=1$, $T_1$ is empty and the equality holds trivially with $k=1$.

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.

Conjugacy classes. For $a\in G$, its conjugacy class isEvery element belongs to its own conjugacy class, by taking $g=e$. To prove uniqueness, suppose $x=ga{g}^{-1}=hb{h}^{-1}$ belongs to both $\mathrm{Cl}(a)$ and $\mathrm{Cl}(b)$. ThenFor every $t\in G$, this makes $tb{t}^{-1}$ a conjugate of $a$ as well. Hence $\mathrm{Cl}(b)\subseteq \mathrm{Cl}(a)$, and reversing the argument gives the other inclusion. Two conjugacy classes that meet are therefore equal, proving that the classes partition $G$.

Abelian case. If $G$ is an abelian group, then $ga{g}^{-1}=ag{g}^{-1}=a$ for every $g$, so $\boxed{\mathrm{Cl}(a)=\{a\}}$ for each $a$.

Centre. An element has a singleton conjugacy class precisely when $gz{g}^{-1}=z$ for every $g$, equivalently $gz=zg$ for every $g$. Thus the definition in the paper agrees with the usual centre of a group:The identity lies in $Z(G)$. If $z,w\in Z(G)$, then for any $g$, $gzw=zgw=zwg$, so $zw\in Z(G)$. Also $gz=zg$ implies $z^{-1}g=g{z}^{-1}$, so $z^{-1}\in Z(G)$. This proves it is a subgroup. Any two of its elements commute because either one commutes with every element of $G$. Consequently $\boxed{Z(G)\text{ is an abelian subgroup of }G}$.

Square symmetry. Use the paper's notation $D_4$ for the eight-element dihedral group of a square. Let $r$ be a quarter-turn and $s$ a reflection, so $r^4=s^2=e$ and $srs=r^{-1}$. A rotation $r^j$ commutes with $s$ exactly when $r^j=r^{-j}$, or $2j=0$ modulo four. This selects $e$ and $r^2$. No reflection $r^{j}s$ commutes with $r$, becauseand $r^{j-1}\ne r^{j+1}$. Therefore

If $h=ag{a}^{-1}$, the representation law gives $D(h)=D(a)D(g)D(a{)}^{-1}$. Trace invariance under this similarity transformation givesThus the character of a representation is constant on every conjugacy class.

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.

The Klein four-group has multiplication tableSince it is abelian, every conjugacy class is a singleton: $\{1\},\{a\},\{b\},\{c\}$. An irreducible representation is a nonzero representation whose only invariant subspaces are zero and the whole representation space. For a finite group, the number of inequivalent complex irreducible representations equals the number of conjugacy classes, so there are four here.

A one-dimensional representation sends the involutions $a,b$ to signs, independently. All four choices areThey are distinct irreducible representations, exhausting the four possibilities. The character table of the Klein four-group isThe specified diagonal matrices obey $D(a{)}^2=D(b{)}^2=I$, commute, and satisfy $D(a)D(b)=D(c)$, so the group relations prove that $D$ is a group representation. Its coordinate axes are invariant lines carrying respectively ${\chi }_{01},{\chi }_{10},{\chi }_{11}$. ThereforeThe trivial representation ${\chi }_{00}$ is absent. If $H$ denotes the four-row character matrix above, direct row products give $H{H}^T=4I$. This verifies character orthogonality:The character of $D$ is $(3,-1,-1,-1)$, whose inner products with the four rows are $0,1,1,1$, independently confirming the decomposition. All these irreducibles are real as well: commuting real involutions can be simultaneously diagonalized into their real sign eigenspaces.

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.

Two elements are conjugate group elements if $h=xg{x}^{-1}$ for some $x\in G$. The conjugacy class of $g$ is $\{xg{x}^{-1}:x\in G\}$. Conjugacy is an equivalence relation: it is reflexive using $x=e$, symmetric by replacing $x$ with $x^{-1}$, and transitive becauseThus the classes partition the group: every element belongs to its own class, and any two classes that intersect are equal. This proves that every element belongs to exactly one class.

If $G$ is Abelian, then $xg{x}^{-1}=g$, so every class is a singleton. In general the elements whose classes are singletons are precisely those commuting with every group element. Thus the center of a group isIt contains $e$. If $a,b\in Z(G)$, then $a{b}^{-1}$ commutes with every $g$, since $b^{-1}$ does and $a$ does. The subgroup test therefore proves that $Z(G)$ is a subgroup. Any two of its elements commute with each other, so it is an abelian subgroup.

For the square centred at the origin with vertices $(\pm 1,\pm 1)$, putThe dihedral group $D_4$ consists of $R^j$ and $R^{j}S$, $j=0,1,2,3$. Its eight actions and their matrices areThese exhaust the symmetries: the image of one vertex has four choices and the image of an adjacent vertex has two choices, determining the entire square. For an explicit pair of nonconjugate elements take $R^2$ and $S$. Both have order two, but their determinants are $+1$ and $-1$. A similarity transformation preserves the determinant, so $R^2$ and $S$ are not conjugate.

To find the centre, use $SR=R^{-1}S$. A rotation $R^j$ commutes with $S$ exactly when $R^j=R^{-j}$, so $j$ is even. No reflection $R^{j}S$ commutes with $R$, since $(R^{j}S)R=R^{j-1}S$ while $R(R^{j}S)=R^{j+1}S$, and $R^2\ne I$. Therefore

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.