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The Peter-Weyl theorem makes finite-dimensional unitary representations separate elements of a compact group. Intersecting their closed kernels gives the identity subgroup. If the descending chain condition holds for closed subgroups, choose finitely many representations whose kernel intersection is already trivial: otherwise repeatedly exclude a surviving nonidentity element to obtain an infinite strict chain. Their direct sum is a faithful representation into a finite-dimensional unitary group. Compactness makes its image closed, and the closed-subgroup theorem gives the original group a compatible Lie group structure.

Let $H$ be an open normal subgroup of a compact topological group $G$, and let $\tau :H\to {\mathrm{GL}}_m({\mathbb{Z}}_p)$ be a continuous faithful representation. Choose representatives $t_j$ for the finite left cosets $G/H$. Write $g{t}_j=t_i{h}_{ij}$ and act on the $j$-th summand of ${⨁}_j{\mathbb{Z}}_p^m$ by $\tau (h_{ij})$ into the $i$-th summand. The cocycle identity proves this is a continuous group representation. An element outside $H$ has a nontrivial coset permutation, while an element of $H$ acting trivially has $\tau (g)=1$ on the identity-coset summand. Hence the representation of $G$ is faithful and integral.

The permutation matrices of the three displayed double transpositions, together with $I_4$, form a faithful representation of the Klein four-group. They satisfy $P^2=Q^2=I_4$ and $PQ=QP$. Every nonidentity element fixes no basis vector, so this is its four-dimensional regular representation.

The constant vector $(1,1,1,1{)}^T$ spans a nonzero proper invariant subspace, proving reducibility. More completely, the columns ofform a common orthonormal basis of eigenvectors. Direct multiplication gives $S^{T}PS=\mathrm{diag}(1,1,-1,-1)$ and $S^{T}QS=\mathrm{diag}(1,-1,1,-1)$. Thus the representation splits into the four distinct real one-dimensional characters, each once. The basis change is a normalized Hadamard matrix.

Every uniform pro-p group embeds continuously in ${\mathrm{GL}}_n({\mathbb{Z}}_p)$ for some finite $n$. Apply the Ado theorem to ${\mathbb{Q}}_p\otimes L_G$ and choose a deep power lattice whose representing matrices lie in $p\mathrm{End}(\Lambda )$, or $4\mathrm{End}(\Lambda )$ at $p=2$. The matrix exponential and logarithm give a faithful representation of the corresponding open normal power subgroup. Induce that representation to the whole group using its finitely many cosets. An element outside the open subgroup permutes the coset blocks nontrivially; an element inside it is detected by the original representation. Thus the induced representation is faithful. All block matrices preserve a finite free ${\mathbb{Z}}_p$-module, giving the integral embedding.

For a real orthogonal involution, $A^{-1}=A=A^T$, so it is a symmetric matrix. Its eigenvalues are $\pm 1$. In two dimensions the possibilities are the identity, the half-turn $-I$, and reflectionsThe reflection fixes the line making angle $\theta /2$ with the first coordinate axis and negates its perpendicular direction. This follows either from the two corresponding eigenvectors or by rotating $\mathrm{diag}(1,-1)$. A nonidentity element of order two in a faithful representation must be either the half-turn or a reflection; the identity matrix is excluded by faithfulness.

By part (i), finite-dimensional continuous unitary representations separate points of the compact group. More explicitly, if every representation sent some $g\ne e$ to the identity, all their matrix coefficients would take the same values at $g$ and $e$. Uniform density would force every continuous scalar function to do so, contradicting separation of points on a compact Hausdorff space. HenceEach representation kernel is a closed normal subgroup. We claim that the intersection can be reduced to finitely many kernels. Begin with $K_0=G$. If $K_j\ne \{e\}$, choose $g_j\in K_j\setminus \{e\}$ and a representation ${\theta }_{j+1}$ with ${\theta }_{j+1}(g_j)\ne I$. ThenAn indefinite continuation would violate the descending chain condition for closed subgroups. Thus after finitely many steps, $K_N=\{e\}$.

The direct sum $\rho ={\theta }_1\oplus \cdots \oplus {\theta }_N$ is consequently a finite-dimensional faithful representation, with continuous injective homomorphismCompactness of $G$ makes this a homeomorphism onto a compact, hence closed, image. The closed-subgroup theorem makes that image an embedded Lie subgroup of the finite-dimensional unitary group. Transporting its smooth structure to $G$ provesThis is the mechanism behind the assertion that compact groups with a descending chain condition are Lie groups: the chain condition turns an arbitrarily large separating family into one finite-dimensional faithful representation.

A faithful representation of dimension one embeds $A$ into ${\mathbb{C}}^{\times }$. Every image element is a $|A|$th root of unity by Lagrange's theorem, so the image is a subgroup of the cyclic group of those roots. It is cyclic, and faithfulness makes $A$ cyclic. Conversely a generator of a cyclic group of order $m$ can be sent to $e^{2\pi i/m}$, giving an injective one-dimensional character. Thus such a faithful representation exists exactly when $A$ is a cyclic group.

The objective is a continuous faithful representation of a uniform pro-p group $G$ in some ${\mathrm{GL}}_N({\mathbb{Z}}_p)$. The proof proceeds from the intrinsic Lie lattice to a small matrix group, then extends the resulting representation to all of $G$. This last step matters: a representation defined only on an open subgroup does not yet establish linearity of the original group.

Let $L=L_G$ be the Lie algebra of a uniform pro-p group and put $\mathfrak{g}={\mathbb{Q}}_p{\otimes }_{{\mathbb{Z}}_p}L$. The uniform-group/Lie-lattice correspondence identifies the group law on $L$ with BCH. The substantial finite-dimensional Lie-theoretic input is the Ado theorem: every finite-dimensional Lie algebra over a characteristic-zero field admits a faithful finite-dimensional representation. Apply it to obtainAn Adjoint representation would not suffice here, since it kills the centre; Ado supplies injectivity even in central directions.

Choose a p-adic lattice $\Lambda$ spanning $V$. Since $L$ has finitely many lattice generators, their representing matrices have bounded denominators. There is therefore an integer $a\ge 0$ such thatOn this matrix lattice the series $\exp X={\sum }_{n\ge 0}{X}^n/n!$ and $\log (1+Y)={\sum }_{n\ge 1}(-1{)}^{n+1}{Y}^n/n$ converge and are inverse. For example $v_p(n!)\le n/(p-1)$, so the valuations of the exponential terms tend to infinity with this choice of $ϵ$; the logarithm terms do as well. They take values in $1+p^{ϵ}\mathrm{End}(\Lambda )$ and in $p^{ϵ}\mathrm{End}(\Lambda )$ respectively. The formal identity $\exp (\mathrm{BCH}(X,Y))=\exp X\exp Y$ holds for the convergent series.

Under the correspondence, $U=G^{p^a}$ has Lie lattice $p^{a}L$ and is an open characteristic subgroup. Definewhere ${\log }_G$ denotes the identification of $U$ with $p^{a}L$. A Lie homomorphism preserves BCH, so the exponential identity proves $\eta (uv)=\eta (u)\eta (v)$. If $\eta (u)=I$, the matrix logarithm gives $\rho ({\log }_{G}u)=0$; faithfulness of $\rho$ gives $u=1$. Thus $\eta$ is a continuous faithful integral representation of $U$.

To extend it, take the moduleValues on a transversal of $U\backslash G$ determine $f$, so $W$ is a free module over ${\mathbb{Z}}_p$ of rank $[G:U]\dim V$. These functions are continuous, since on each open coset they are given by the continuous representation $\eta$. Define $(T_{g}f)(x)=f(xg)$. The equivariance condition is preserved, and $T_g{T}_h=T_{gh}$. In a transversal basis each $T_g$ is a permutation of blocks followed by matrices from $\eta (U)$, so it belongs to $\mathrm{GL}(W)$ and depends continuously on $g$.

Faithfulness can be checked directly. If $g\notin U$, right multiplication moves the coset $U$ to a different coset, so $T_g$ moves a function supported on one block and cannot be the identity. If $g\in U$ and $T_g=1$, evaluation at $1$ gives $f(g)=f(1)$ for every $f$, whereas equivariance gives $f(g)=\eta (g)f(1)$. Since $f(1)$ is arbitrary, $\eta (g)=I$, and hence $g=1$. ThereforeCompactness of $G$ makes this continuous injection a homeomorphism onto its closed image. This proves the linearity of a uniform pro-p group for the full group, including its centre, and explains why taking a sufficiently small power lattice permits matrix integration before finite-index induction restores the whole group.

The first alternative in the supplied analytic-group theorem would give a faithful representation of $\mathcal{N}$ in some ${\mathrm{GL}}_d({\mathbb{Z}}_p)$ or ${\mathrm{GL}}_d({\mathbb{F}}_p[[t]])$. The embedding proved above supplies cyclic subgroups of order $p^a$ for arbitrarily large $a$. Such orders are impossible in either fixed matrix dimension.

In characteristic zero, a matrix $A$ of order $p^a$ is diagonalizable over an algebraic closure, because its minimal polynomial divides the separable polynomial $X^{p^a}-1$. At least one eigenvalue must be a primitive $p^a$th root of unity; otherwise $A^{p^{a-1}}=I$. The cyclotomic polynomial ${\Phi }_{p^a}$ is irreducible over ${\mathbb{Q}}_p$, since ${\Phi }_{p^a}(1+X)$ satisfies the Eisenstein criterion at $p$. Consequently that eigenvalue has degree $\phi (p^a)=p^{a-1}(p-1)$ over ${\mathbb{Q}}_p$, which must be at most the matrix dimension $d$. This bounds $a$.

In characteristic $p$, pass from the power-series ring to its fraction field. A matrix of $p$-power order satisfies $X^{p^a}-1=(X-1{)}^{p^a}$, so it is unipotent. Writing $A=I+B$, nilpotence gives $B^d=0$. For $p^e\ge d$,Thus its order is at most $p^{\lceil {\log }_{p}d\rceil }$, again a fixed bound. Therefore the linear alternative is impossible in both coefficient rings.

For a complex irreducible representation, every central element acts as a scalar by Schur lemma. If the representation is faithful, these scalars embed $Z$ in ${\mathbb{C}}^{\ast }$. Every finite subgroup of ${\mathbb{C}}^{\ast }$ is cyclic: if its order is $m$, all its elements satisfy $z^m=1$, and the $m$th roots of unity form a cyclic group. Thus $Z$ is cyclic.

The trivial group satisfies the conclusion immediately. Otherwise let $G$ be a finite $p$-group with cyclic centre, and decompose a faithful representation as a direct sum of irreducibles using Maschke's theorem. The nontrivial cyclic $p$-group $Z$ has a unique subgroup $C$ of order $p$. Some component does not kill $C$, since otherwise the whole representation would kill it. If that component had a nontrivial kernel $N$, the kernel would be a normal subgroup and the allowed intersection theorem gives $N\cap Z\ne 1$. Every nontrivial subgroup of cyclic $Z$ contains $C$, contradicting the choice of component. That irreducible component is faithful.

Finally the regular representation is faithful, since left multiplication by a nonidentity element does not fix the basis vector at the identity. A cyclic centre therefore supplies a faithful irreducible component of it. The first argument proves the converse, giving a finite $p$-group has a faithful irreducible complex representation exactly when its centre is cyclic.

If $\rho$ is a faithful representation, $g\mapsto {\lambda }_g$ embeds $Z$ into the multiplicative group of complex numbers. Its image is a finite group of roots of unity, hence lies in the cyclic group of $n$th roots of unity and is itself cyclic. Thus $Z$ is cyclic.

The complex regular representation has basis $(e_h{)}_{h\in G}$ and left actionIf $\rho (g)$ is the identity, thenso $g=1$. Thus $\mathbb{C}G$ is a faithful representation.

By Maschke's theorem, the regular representation decomposes as a direct sum of irreducibles,Its kernel is the intersection of the kernels of the constituent representations:Each constituent kernel is a normal subgroup of $G$. If $G$ is simple, every such kernel is either $\{1\}$ or $G$. They cannot all be $G$, because then their intersection would be $G$, contradicting faithfulness of the regular representation. Hence some $V_j$ has trivial kernel. Therefore every finite simple group has a faithful irreducible representation of a finite simple group.

Schur lemma says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A continuous representation of a topological group is a continuous homomorphismfor a finite-dimensional complex vector space $V$. It is a unitary representation if $V$ has a positive-definite Hermitian inner product for whichfor every $g\in G$ and $v,w\in V$.

For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized Haar measure:Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving unitarity. Since $S^1$ is abelian, the operators $\rho (z)$ commute; since they are unitary, they are normal. Simultaneous diagonalization therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every representation of the circle group is a direct sum of one-dimensional representations.

WriteThe group law and inverse areA calculation givesEvery element of the centre $Z$ occurs by taking, for example, $x^{\prime }=y=0$ and $y^{\prime }=1$. Hence the commutator subgroup is exactly $Z$. The image of a one-dimensional representation is abelian, so the one-dimensional representation kills the commutator subgroup and its kernel contains $Z$.

Now let $(\rho ,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroupis a representation of the circle group. Decompose it into its distinct weight spaces:Because $Z/Z_0$ is central, every $\rho (g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the central circle weight-space decomposition.

Let $n_i=\dim V_i$. The mapis a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value isso ${\theta }_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every ${\theta }_i$ is trivial. Since the ${\theta }_i$ were distinct, $d=1$ and ${\theta }_1=1$.

It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is faithful. This is the real Heisenberg quotient has no faithful finite-dimensional representation.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$ into invertible linear maps. It is a faithful representation when it is injective, equivalently when its kernel contains only the identity. A faithful real two-dimensional representation of the cyclic group of order four isThese are rotations of the plane through $0,\pi /2,\pi ,3\pi /2$. Since $J^2=-I$ and $J^4=I$, their multiplication agrees with the group, and all four matrices are distinct.

For any invertible $S$, set $\tilde{D}(g)=SD(g)S^{-1}$. Thenso this also is a group representation. The character of a representation is ${\chi }_D(g)=\mathrm{tr}D(g)$. For the displayed quarter-turn representation, its values on $1,i,-1,-i$ are $2,0,-2,0$.

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.

Let $B=\left(\begin{array}{cc}0& b\\ c& 0\end{array}\right)$ be the lower block. ThenThe defining relation $a^4=I$ requires $T(a{)}^4=I_3$, so for real constants $bc=\pm 1$. Both possibilities ensure invertibility.

If $bc=-1$, the generator matrix has order four, because its square is not the identity while its fourth power is. Its four powers are distinct, giving a faithful representation. If $bc=1$, its square is the identity but it is not itself the identity; its lower diagonal entries are zero. The image then has order two and the kernel of a group homomorphism is $\{I,a^2\}$. Thus the faithfulness criterion for a block representation of the cyclic group of order four givesOther parameter values give no representation of $Z_4$, not additional unfaithful cases; $bc=0$ also makes the assigned matrix singular. No orthogonality of these matrices was required.

For the final construction takeMultiplication adds exponents, so $D(a^r)D(a^s)=D(a^{r+s})$ and $D(a{)}^4=I_2$. The representing matrix is the identity exactly for even exponents. Therefore this is a two-dimensional representation with kernel $\{I,a^2\}$, as required. It factors through the quotient cyclic group of order two.

Denote the four listed matrices, in order, by $I,P,Q,W$. As permutation matrices, $P,Q,W$ act on the four basis vectors by $(12)(34)$, $(13)(24)$ and $(14)(23)$, respectively. Direct multiplication givesThese relations prove closure and show that each element is its own inverse. The identity is included and matrix multiplication is associative, so the matrices form a group.

The relations coincide with those of the Klein four-group $D_2$: take its half-turn generator to $P$, a reflection generator to $Q$, and their product to $W$. All four matrices are distinct, so this assignment is a faithful representation of $D_2$. It is the four-point regular representation of the Klein four-group.

Every matrix fixes $v=(1,1,1,1{)}^T$, since it permutes its coordinates. Thereforeis a one-dimensional invariant subspace, nonzero and strictly smaller than the four-dimensional space. This proves the representation is reducible. In fact the four mutually orthogonal sign vectors $(1,1,1,1)$, $(1,1,-1,-1)$, $(1,-1,1,-1)$ and $(1,-1,-1,1)$ give invariant lines and completely diagonalize the representation, as in the normalized Hadamard matrix construction.

For a group representation $D:G\to \mathrm{GL}(V)$, a faithful representation is injective, equivalently $D(g)=I$ only for $g=e$. Thus no distinct group elements are represented by the same linear map.

Two equivalent representations are related by one fixed invertible change of basis: $E(g)=S^{-1}D(g)S$ for every $g$. The same $S$ must work simultaneously for all group elements.

An irreducible representation acts on a nonzero vector space and has no invariant subspace other than $\{0\}$ and the whole space. An invariant subspace is one carried into itself by every representing matrix.

The character of a representation isThe cyclic property of the trace makes it invariant under equivalence and constant on conjugacy classes. Its value at the identity is the dimension of the representation.

A group representation assigns each $g\in G$ an invertible linear map $D(g)$ on a fixed vector space, preserving multiplication:Choosing a basis turns those maps into matrices. A faithful representation is injective, equivalently its kernel is $\{e\}$, so distinct elements have distinct matrices.

Two group representations on spaces of the same dimension are called equivalent representations if one fixed invertible change of basis conjugates every representing matrix: $E(g)=S^{-1}D(g)S$ for all $g$. Choosing a different $S$ separately for each element does not establish equivalence.

The character of a representation isThe cyclic property of the trace makes it invariant under a change of basis and constant on conjugacy classes. Thus different characters at even one group element prove inequivalence.