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Conjugation by $P$ partitions $H$ into orbits. Each non-singleton orbit has size divisible by $p$, by the orbit-stabilizer theorem. The singleton orbits are exactly $H\cap Z(P)$. Since $|H|$ is divisible by $p$, $|H\cap Z(P)|\equiv 0(\mathrm{mod}p)$. This set contains $1$, so $\boxed{|H\cap Z(P)|\ge p}$.

For the deduction we also give the representation-theoretic ingredient behind Burnside's theorem: a nonabelian finite simple group has no nontrivial conjugacy class of prime power size. Here is a proof, so no stronger theorem is being assumed without justification. Suppose $g\ne 1$ has class size $p^b$. For an irreducible character $\chi$ of degree $d$ coprime to $p$, the class sum acts as the scalar $p^b\chi (g)/d$, which is an algebraic integer: the class sum has an integer-entry matrix in the regular representation, which contains every irreducible representation, so this scalar is an eigenvalue of an integer-entry matrix. Also $\chi (g)$ is an algebraic integer. Bezout's identity then makes $\chi (g)/d$ an algebraic integer. All its algebraic conjugates have modulus at most $1$, because character values of a representation are sums of $d$ roots of unity. An algebraic integer with this property is zero or a root of unity: the monic polynomials of all its powers have uniformly bounded integer coefficients, so two powers must coincide unless it is zero.

Every nontrivial irreducible representation of a simple group is faithful. If $\chi (g)/d$ were a root of unity, equality in the triangle inequality would force its representing matrix to be scalar, hence $g$ central, impossible. Thus $\chi (g)=0$ whenever $p\nmid d$, except for the trivial character of a representation. Orthogonality of the identity column and the $g$ column in the character table now givesThis would make $1/p$ an algebraic integer, contradicting the fact that rational algebraic integers are integers. The same proof covers class size $1$ directly through triviality of the center.

If an abelian subgroup $A$ had index $p^a$, every $g\in A\setminus \{1\}$ would have $A\subseteq C_G(g)$, making its class size a divisor of $p^a$, a contradiction. Such a $g$ exists: if $A=1$, then $G$ is a nontrivial p-group, whose center is nontrivial by the first argument with $H=P$. Therefore a nonabelian simple group has no abelian subgroup of prime power index.

Multiplicativity of the determinant gives $\delta (gh)=\delta (g)\delta (h)$ and $\delta (1)=1$, so $\delta$ is a linear character. Its finite image is a cyclic group of roots of unity. If it contains $-1$, its order is $2m$ and the character of a representation ${\delta }^m$ maps onto $\{\pm 1\}$. The kernel of a group homomorphism is the preimage of its identity, so $\ker ({\delta }^m)$ is a normal subgroup of index $2$. The subgroup $\ker \delta$ itself need not have index $2$.

By Cauchy theorem for groups, a group $E$ of order $2k$ has an involution $g$. In the regular representation, left multiplication by $g$ is a permutation consisting of $k$ disjoint transpositions, so its determinant is $(-1{)}^k=-1$ when $k$ is odd. The preceding argument gives a normal subgroup of index two.

For a nonabelian simple group of order below $80$, Burnside's theorem requires at least three distinct prime factors. An odd order with three factors is at least $3\cdot 5\cdot 7=105$. An even order with exactly one factor of $2$ has a normal subgroup of index $2$. Thus its order must be divisible by $4$ and by at least two odd primes. Below $80$ the only possibility is $4\cdot 3\cdot 5=60$, since the next is $4\cdot 3\cdot 7=84$. Hence $\boxed{|H|=60}$.

The row orthogonality relations for a character table state that for irreducible characters $\chi ,\psi$,Equivalently, if $c$ runs through representatives of the conjugacy classes $C$,

Fix an irreducible $\chi$. The sum of the elements in a conjugacy class $C$ is central in $\mathbb{C}G$, so Schur lemma says that it acts in the representation affording $\chi$ by the scalarThis central character value of a conjugacy-class sum is an algebraic integer: the class sum acts by a matrix with integer entries on the regular representation, and ${\omega }_{\chi }(C)$ is one of its eigenvalues. Also $\overline{\chi (c)}$ is an algebraic integer because character values are sums of roots of unity.

Row orthogonality with $\psi =\chi$ now givesThe right-hand side is an algebraic integer. The left-hand side is rational, and every rational algebraic integer is an integer. Therefore

Write the finite abelian group additively. Its group shift operator isThese unitary operators form the regular representation and commute. The representation-theoretic facts we use are that every irreducible representation of a finite abelian group is one-dimensional, there are $|G|$ such characters of a representation, and their character orthogonality gives an orthonormal basis of functions on $G$. Each linear character here is a group homomorphism $\chi :G\to U(1)$, with $|\chi (g)|=1$.

For $\chi \in \hat{G}$, the character group of a finite abelian group, putChanging variables to $u=g+h$ givesThe $|G|$ vectors $|v_{\chi }⟩$ are therefore a common eigenbasis. Using $\chi (g)$ instead of its complex conjugate in their definition reverses every eigenphase; this is merely the quantum Fourier transform sign convention.

Let $G=C_3=⟨g⟩$ and let $e_0,e_1,e_2$ be the standard basis of its real regular representation, with $g{e}_j=e_{j+1}$ modulo three. SetBoth are invariant, they are orthogonal, andThe line $U$ is the trivial representation. On the plane $W$, the generator acts as a rotation through $2\pi /3$. Such a rotation has no real eigenline, so $W$ has no nonzero proper invariant real subspace and is irreducible.

Let ${\chi }_{\mathrm{reg}}$ be the character of the regular representation. It has ${\chi }_{\mathrm{reg}}(1)=|G|$ and vanishes away from the identity. The multiplicity of an irreducible character ${\chi }_j$ in it isThus the regular representation decomposes as ${⨁}_j{n}_j{\rho }_j$. Comparing dimensions gives

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.

A complex representation of $G$ is a complex vector space $V$ together with a homomorphismIt is faithful when $\ker \rho =\{e\}$, equivalently when $\rho$ is injective. Every finite group has a faithful complex representation: in the regular representation, $G$ permutes the basis $\{e_h:h\in G\}$ byand a group element fixing every basis vector must be the identity.

If $g$ is conjugate to $g^k$, then $\rho (g)$ is similar to $\rho (g^k)=\rho (g{)}^k$. Thus the two matrices have the same spectrum, while an eigenvalue $\lambda$ of $\rho (g)$ gives the eigenvalue ${\lambda }^k$ of $\rho (g{)}^k$. Hence

Now take a $p$-cycle $g\in S_p$. It is conjugate to $g^k$ for every $1\le k<p$. Faithfulness makes $\rho (g)$ have order $p$, so its spectrum contains a nontrivial $p$th root of unity $\lambda$. The preceding implication puts all the $p-1$ distinct valuesin the spectrum. Therefore $\dim V\ge p-1$. This is the spectrum orbit bound for a faithful symmetric-group representation.

Finally let $G$ be the group of all permutations of $\mathbb{N}$. It has no faithful finite-dimensional complex representation. For every $n$, $G$ contains a subgroup $(C_2{)}^n$ generated by $n$ disjoint transpositions. In a $d$-dimensional complex representation, commuting involutions are simultaneously diagonalizable and hence map into the group of diagonal sign matrices, which has order $2^d$. A faithful restriction would require $2^n\le 2^d$ for every $n$, an impossibility.

If ${\rho }_1,\dots ,{\rho }_k$ are the complex irreducible representations of a finite group $G$ and $n_j=\dim {\rho }_j$, decomposition of the regular representation gives