hcju/proofwikiqa huggingface dataset:
Let $G$ be any arbitrary finite group whose identity is $e_G$. Let $S$ be the symmetric group on the elements of $G$, where $e_S$ is the identity of $S$. For any $x \in G$, let $\lambda_x$ be the left regular representation of $G$ with respect to $x$. From Regular Representations in Group are Permutations, $\forall x \in G: \lambda_x \in S$. So, we can define a mapping $\theta: G \to S$ as: :$\forall x \in G: \map \theta x = \lambda_x$ From Composition of Regular Representations, we have: :$\forall x, y \in G: \lambda_x \circ \lambda_y = \lambda_{x y}$ where in this context $\circ$ denotes composition of mappings. Thus by definition of $\theta$: :$\map \theta x \circ \map \theta y = \map \theta {x y}$ demonstrating that $\theta$ is a homomorphism. Having established that fact, we can now consider $\map \ker \theta$, where $\ker$ denotes the kernel of $\theta$. Let $x \in G$. We have that: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map \ker \theta       | c =  }} {{eqn | ll= \leadsto       | l = \map \theta x       | r = e_S       | c = {{Defof|Kernel of Group Homomorphism}} }} {{eqn | ll= \leadsto       | l = \lambda_x       | r = I_G       | c = where $I_G$ is the identity mapping of $G$ }} {{eqn | ll= \leadsto       | l = \map {\lambda_x} {e_G}       | r = \map {I_G} {e_G}       | c = {{Defof|Identity Mapping}} }} {{eqn | ll= \leadsto       | l = x       | r = e_G       | c = Definitions of $\lambda_x$ and $I_G$ }} {{end-eqn}} So $\map \ker \theta$ can contain no element other than $e_G$. So, since clearly $e_G \in \map \ker \theta$, it follows that: :$\map \ker \theta = \set {e_G}$ By Kernel is Trivial iff Monomorphism, it follows that $\theta$ is a monomorphism. By Monomorphism Image is Isomorphic to Domain, we have that: :$\theta \sqbrk G \cong \Img \theta$ that is, $\theta$ is isomorphic to its image. Hence the result. {{qed}}
If $\left({G, +_G: \circ}\right)_R$ is an $R$-module, then: $\forall x, y, \in G, \forall \lambda, \mu \in R$: : $(1): \quad \lambda \circ \left({x +_G y}\right) = \left({\lambda \circ x}\right) +_G \left({\lambda \circ y}\right)$ : $(2): \quad \left({\lambda +_R \mu}\right) \circ x = \left({\lambda \circ x}\right) +_G \left({\mu \circ x}\right)$ : $(3): \quad \left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$ If $\phi: G \to H$ is an epimorphism, then: : $\forall x, y \in G: \phi \left({x +_G y}\right) = \phi \left({x}\right) +_H \phi \left({y}\right)$ : $\forall x \in S: \forall \lambda \in R: \phi \left({\lambda \circ x}\right) = \lambda \circ \phi \left({x}\right)$ : $\forall y \in H: \exists x \in G: y = \phi \left({x}\right)$ As $\phi$ is an epimorphism, we can accurately specify the behaviour of all elements of $H$, as they are the images of elements of $G$. If $\phi$ were not an epimorphism, i.e. not surjective, we would have no way of knowing the behaviour of elements of $H$ outside of the image of $G$. Hence the specification that $\phi$ needs to be an epimorphism. Now we check the criteria for $H$ being a module, in turn. === Module $(1)$ === {{begin-eqn}} {{eqn | l=\lambda \circ \left({\phi \left({x}\right) +_H \phi \left({y}\right)}\right)       | r=\lambda \circ \left({\phi \left({x +_G y}\right)}\right)       | c= }} {{eqn | r=\phi \left({\lambda \circ \left({x +_G y}\right)}\right)       | c= }} {{eqn | r=\phi \left({\left({\lambda \circ x}\right) +_G \left({\lambda \circ y}\right)}\right)       | c= }} {{eqn | r=\phi \left({\lambda \circ x}\right) +_H \phi \left({\lambda \circ y}\right)       | c= }} {{end-eqn}} Thus $(1)$ is shown to hold for $H$. === Module $(2)$ === {{begin-eqn}} {{eqn | l=\left({\lambda +_R \mu}\right) \circ \phi \left({x}\right)       | r=\phi \left({\left({\lambda +_R \mu}\right) \circ x}\right)       | c= }} {{eqn | r=\phi \left({\left({\lambda \circ x}\right) +_G \left({\mu \circ x}\right)}\right)       | c= }} {{eqn | r=\phi \left({\lambda \circ x}\right) +_H \phi \left({\mu \circ x}\right)       | c= }} {{eqn | r=\lambda \circ \phi \left({x}\right) +_H \mu \circ \phi \left({x}\right)       | c= }} {{end-eqn}} Thus $(2)$ is shown to hold for $H$. === Module $(3)$ === {{begin-eqn}} {{eqn | l=\left({\lambda \times_R \mu}\right) \circ \phi \left({x}\right)       | r=\phi \left({\left({\lambda \times_R \mu}\right) \circ x}\right)       | c= }} {{eqn | r=\phi \left({\lambda \circ \left({\mu \circ x}\right)}\right)       | c= }} {{eqn | r=\lambda \circ \left({\phi \left({\mu \circ x}\right)}\right)       | c= }} {{eqn | r=\lambda \circ \left({\mu \circ \phi \left({x}\right)}\right)       | c= }} {{end-eqn}} Thus $(3)$ is shown to hold for $H$. So all the criteria for $H$ to be an $R$-module are fulfilled. {{qed}}
Any linear transformation clearly satisfies the condition. Let $\phi$ be such that the condition is satisfied. Let $\lambda = \mu = 1_R$. Then $\map \phi {x + y} = \map \phi x + \map \phi y$. Now let $\mu = 0_R$. Then $\map \phi {\lambda x} = \lambda \map \phi x$. Thus the conditions are fulfilled for $\phi$ to be a homomorphism, that is, a linear transformation. {{Qed}}
To show that $\struct {G^S, +_G', \circ}_R$ is an $R$-module, we verify the following: $\forall f, g, \in G^S, \forall \lambda, \mu \in R$: :$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$ :$(2): \quad \paren {\lambda +_R \mu} \circ f = \paren {\lambda \circ f} +_G \paren {\mu \circ f}$ :$(3): \quad \paren {\lambda \times_R \mu} \circ f = \lambda \circ \paren {\mu \circ f}$ === Criterion 1 === :$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$ Let $x \in S$. Then: {{begin-eqn}} {{eqn | l = \lambda \circ \paren {\map {\paren {f +_G' g} } x}       | r = \lambda \circ \paren {\map f x +_G \map g x}       | c =  }} {{eqn | r = \paren {\lambda \circ \map f x} +_G \paren {\lambda \circ \map g x}       | c = {{Defof|Module}} }} {{eqn | r = \paren {\map {\paren {\lambda \circ f} } x} +_G \paren {\map {\paren {\lambda \circ g} } x}       | c =  }} {{eqn | r = \map {\paren {\paren {\lambda \circ f} +_G' \paren {\lambda \circ g} } } x       | c =  }} {{end-eqn}} Thus $(1)$ holds. {{qed|lemma}} === Criterion 2 === : $(2): \quad \paren {\lambda +_R \mu} \circ f = \paren {\lambda \circ f} +_G \paren {\mu \circ f}$ Let $x \in S$. {{begin-eqn}} {{eqn | l = \map {\paren {\paren {\lambda +_R \mu} \circ f} } x       | r = \paren {\lambda +_R \mu} \circ \paren {\map f x}       | c =  }} {{eqn | r = \paren {\lambda \circ \map f x} +_G \paren {\mu \circ \map f x}       | c =  }} {{eqn | r = \map {\paren {\lambda \circ f} } x +_G \map {\paren {\mu \circ f} } x       | c =  }} {{end-eqn}} Thus $(2)$ holds. {{qed|lemma}} === Criterion 3 === :$(3): \quad \paren {\lambda \times_R \mu} \circ f = \lambda \circ \paren {\mu \circ f}$ {{begin-eqn}} {{eqn | l = \map {\paren {\paren {\lambda \times_R \mu} \circ f} } x       | r = \paren {\lambda \times_R \mu} \circ \paren {\map f x}       | c =  }} {{eqn | r = \lambda \circ \paren {\mu \circ \map f x}       | c =  }} {{eqn | r = \lambda \circ \paren {\map {\paren {\mu \circ f} } x}       | c =  }} {{eqn | r = \map {\paren {\lambda \circ \paren {\mu \circ f} } } x       | c =  }} {{end-eqn}} Thus $(3)$ holds. {{qed|lemma}} Hence the result. {{qed}}
Let $\Gamma$ be a graph with $V$ vertices, $E$ edges and $F$ faces. The proof proceeds in two parts: $(1) \implies (2)$ and $(2) \implies (1)$. === $(1)$ implies $(2)$ === First we consider $K_5$.   $K_5$ has $5$ vertices and $10$ edges. By the Euler Polyhedron Formula, a planar embedding would have $7$ faces.   But each face has at least $3$ edges, while each edge bounds at most two faces. Suppose we count the incident edge-face pairs. Then the number of faces is at most: :$E \dfrac 2 3 = 6 \dfrac 2 3$ which contradicts the deduction that $K_5$, if planar, should have $7$ faces. Hence $K_5$ is non-planar. Now consider $K_{3, 3}$. $K_{3, 3}$ has $6$ vertices and $9$ edges. Hence $K_{3, 3}$ has $5$ faces if it is planar. Since $K_{3, 3}$ is bipartite, from Graph is Bipartite iff No Odd Cycles, each cycle in $K_{3,3}$ has an even number of edges. Hence any face must have at least $4$ edges. So the number of faces is at most :$E \dfrac 2 4 = 4 \dfrac 1 2$ which contradicts the deduction that $K_{3, 3}$, if planar, should have $5$ faces. Hence $K_{3, 3}$ is non-planar. Now consider a subdivision of either graph. <!-- A subdivision of a graph is a replacement of any edge $[x,y]$ with two edges, $[x,z]$ and $[z,y]$, where $z$ is a new vertex. --> The arguments as above continue to work as before, since both $V$ and $E$ have increased by one in the Euler formula. So the expected value of $F$ is unchanged. Hence if a graph $\Gamma$ contains a subdivision of $K_5$ or $K_{3,3}$, then $\Gamma$ is not planar. Taking the contrapositive: :$\Gamma$ is a planar graph $\implies \Gamma$ does not contain a subdivision of either $K_5$ or $K_{3, 3}$. === $(2)$ implies $(1)$ === {{ProofWanted}} {{namedfor|Kazimierz Kuratowski|cat = Kuratowski}} Category:Graph Theory gedzpex6s6czckd0wgmg0gqzs6smt6d
Let $G$ be a finite abelian group. By means of Abelian Group is Product of Prime-power Order Groups, we factor it uniquely into groups of prime-power order. Then, Abelian Group of Prime-power Order is Product of Cyclic Groups applies to each of these factors. Hence we conclude $G$ factors into prime-power order cyclic groups. The factorisation of $G$ into prime-power order factors is already unique. Therefore, a demonstration of the uniqueness of the secondary factorisation suffices: Suppose $\order G = p^k$ with $p$ a prime. Let $G$ have the following two factorisations: :$G = H_1 \times H_2 \times \cdots \times H_m = K_1 \times K_2 \times \dotsb \times K_n$ where the $H_i$'s and $K_i$'s are non-trivial cyclic subgroups with :$\order {H_1} \ge \order {H_2} \ge \dotsb \ge \order {H_m}|$ :$\order {K_1} \ge \order {K_2} \ge \dotsb \ge \order {K_n}$ We proceed through complete induction on $k$. === Basis for the induction === For $k = 1$, the statement follows from Prime Group is Cyclic. === Induction Hypothesis === Now we assume the theorem is true for all abelian groups of order $p^l$, where $l < k$. === Induction Step === By Power of Elements is Subgroup, $G^p = \set {x^p: x \in G}$ is a proper subgroup of $G$. It follows that: :$G^p = H_1^p \times \cdots \times H_{m'}^p = K_1^p \times \cdots \times K_{n'}^p$ where: :$m'$ is the largest integer $i$ such that $\order {H_i} > p$ :$n'$ is the largest integer $j$ such that $\order {K_j} > p$ This $m'$ and $n'$ ensure that the direct products above do not have trivial factors. Also, by Cauchy's Group Theorem, we have: :$\order {G^p} < \order G$ This means that we can apply the induction hypothesis. It follows that: :$m' = n'$ and: :$\order {H_i^p} = \order {K_i^p}$ for $i = 1, 2, \ldots, m'$. We know that $\order {H_i} = p \order {H_i^p}$, and the same for $K_i$ and $K_i^p$. It follows that $\order {H_i} = \order {K_i}$ for all $i = 1, 2, \ldots, m'$.   Now by construction of $m'$: :$i > m' \implies \order {H_i} = p = \order {K_i}$ It follows that we have: :$\order {H_1} \order {H_2} \dotsm \order {H_{m'} } p^{m - m'} = \order G = \order {K_1} \order {K_2} \dotsm \order {K_{n'} } p^{n - n'}$ Therefore: :$m - m' = n - n'$ and as $m' = n'$, it must follow that $m = n$. The result follows by induction. {{qed}}
We have that: :$\forall a, b \in S: a +_S b = a + b$ :$\forall a, b \in S: a \times_S b = a \times b$ :$\forall a \in S: \forall x \in G = a \circ_S x = a \circ x$ as $+_S$, $\times_S$ and $\circ_S$ are restrictions. Let us verify the module axioms. === {{Module-axiom|1}} === We need to show that: :$\forall a \in S: \forall x, y \in G: a \circ_S \paren {x +_G y} = a \circ_S x +_G a \circ_S y$ We have: {{begin-eqn}} {{eqn | l = a \circ_S \paren {x +_G y}       | r = a \circ \paren {x +_G y} }} {{eqn | r = a \circ x +_G a \circ y       | c = {{Module-axiom|1}} on $\struct {G, +_G, \circ}_R$ }} {{eqn | r = a \circ_S x +_G a \circ_S y }} {{end-eqn}} {{qed|lemma}} === {{Module-axiom|2}} === We need to show that: :$\forall a,b \in S: \forall x \in G: \paren {a +_S b} \circ_S x = a \circ_S x +_G b \circ_S y$ We have: {{begin-eqn}} {{eqn | l = \paren {a +_S b} \circ_S x       | r = \paren {a + b} \circ x }} {{eqn | r = a \circ x + b \circ x       | c = {{Module-axiom|2}} on $\struct {G, +_G, \circ}_R$ }} {{eqn | r = a \circ_S x +_G b \circ_S x }} {{end-eqn}} {{qed|lemma}} === {{Module-axiom|3}} === We need to show that: :$\forall a, b \in S: \forall x \in G: \paren {a \times_S b} \circ_S x = a \circ_S \paren {b \circ_S x}$ We have: {{begin-eqn}} {{eqn | l = \paren {a \times_S b} \circ_S x       | r = \paren {a \times b} \circ x }} {{eqn | r = a \circ \paren {b \circ x}       | c = {{Module-axiom|3}} on $\struct {G, +_G, \circ}_R$ }} {{eqn | r = a \circ_S \paren {b \circ_S x} }} {{end-eqn}} {{qed|lemma}} Thus $\struct {G, +_G, \circ_S}_S$ is an $S$-module. {{qed|lemma}} It remains to prove the final statement: If $\struct {G, +_G, \circ}_R$ is a unitary $R$-module and $1_R \in S$, then $\struct {G, +_G, \circ_S}_S$ is also unitary. To show that $\struct {G, +_G, \circ_S}_S$ is unitary, we must prove that: :$\forall x \in G: 1_R \circ_S x = x$ Since $1_R \in S$ by assumption, the product $1_R \circ_S x$ is defined.  We now have: {{begin-eqn}} {{eqn | l = 1_R \circ_S x       | r = 1_R \circ x }} {{eqn | r = x       | c = Unitary Module Axiom $(4)$ on $\struct {G, +_G, \circ}_R$ }} {{end-eqn}} {{qed}}
The notation $*^n x$ can be written as $x^n$. Let us verify that $\struct {G, *, \circ}_\Z$ is a unitary $\Z$-module by verifying the axioms in turn. === Axiom $(\text M 1)$ === We need to show that $n \circ \paren {x * y} = \paren {n \circ x} * \paren {n \circ y}$. From the definition, $n \circ x = x^n$ and so $n \circ \paren {x * y} = \paren {x * y}^n$ From Power of Product in Abelian Group: :$\paren {x * y}^n = x^n * y^n = \paren {n \circ x} * \paren {n \circ y}$ {{qed|lemma}} === Axiom $(\text M 2)$ === We need to show that $\paren {n + m} \circ x = \paren {n \circ x} * \paren {m \circ x}$. That is, that $x^{n + m} = x^n * x^m$. This is an instance of Powers of Group Elements: Sum of Indices. {{qed|lemma}} === Axiom $(\text M 3)$ === We need to show that $\paren {n \times m} \circ x = n \circ \paren {m \circ x}$. That is, that $x^{n m} = \paren {x^m}^n$. This follows directly from Powers of Group Elements: Product of Indices. {{qed|lemma}} === Axiom $(\text M 4)$ === We need to show that $\forall x \in G: 1 \circ x = x$. That is, that $x^1 = x$. This follows from the definition of Power of Group Element. {{qed|lemma}} Having verified all four axioms, we have shown that $\struct {G, *, \circ}_\Z$ is a unitary $\Z$-module. {{qed}}
$G$ can be seen as functions: :$\displaystyle f: A \to \bigcup_{a \mathop \in A} G_a$ {{explain|Clarify the above sentence.}} Let $a \in A$. Let $x, y \in G_a$. Let $r \in R$. So both $x + y \in G_a$ and $r x \in G_a$. Let $b \in A$. === Case 1 === {{explain|Explain and clarify the notation}} Let $b = a$. Then: :$\map {\map {\inj_a} {x + y} } b = x + y = \map {\map {\inj_a} x} b + \map {\map {\inj_a} y} b$ and: :$\map {\map {\inj_a} {r x} } b = r x = r \, \map {\map {\inj_a} x} b$ {{qed|lemma}} === Case 2 === Let $b \ne a$. Then: :$\map {\map {\inj_a} {x + y} } b = 0 + 0 = \map {\map {\inj_a} x} b + \map {\map {\inj_a} y} b$ and: :$\map {\map {\inj_a} {r x} } b = 0 = \map r 0 = r \, \map {\map {\inj_a} x} b$ Therefore: :$\map {\inj_a} {x + y} = \map {\inj_a} x + \map {\inj_a} y$ and: :$\map {\inj_a} {r x} = r \, \map {\inj_a} x$ {{qed|lemma}} So $\inj_a$ is a homomorphism. Combined with Canonical Injection is Injection gives that $\inj_a$ is a monomorphism. {{qed}}
If the conditions are fulfilled, then: :$x \in H \implies -x = \left({- 1_R}\right) \circ x \in H$ Thus $H$ is a subgroup of $\left({G, +}\right)$ by the Two-Step Subgroup Test, and hence a submodule. {{qed}}
Follows directly from the fact that the trivial subgroup is a subgroup of $\left({G, +_G}\right)$. {{qed}}
Let $\order G = k p^n$ such that $p \nmid k$. Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements. Let $N = \order {\mathbb S}$. Now $N$ is the number of ways $p^n$ elements can be chosen from a set containing $p^n k$ elements.   From Cardinality of Set of Subsets, this is given by: :$N = \dbinom {p^n k} {p^n} = \dfrac {\paren {p^n k} \paren {p^n k - 1} \cdots \paren {p^n k - i} \cdots \paren {p^n k - p^n + 1} } {\paren {p^n} \paren {p^n - 1} \cdots \paren {p^n - i} \cdots \paren 1}$ From Binomial Coefficient involving Power of Prime: :$\dbinom {p^n k} {p^n} \equiv k \pmod p$ Thus: :$N \equiv k \pmod p$ Now let $G$ act on $\mathbb S$ by the rule: :$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$ That is, $g * S$ is the left coset of $S$ by $g$. From Group Action on Sets with k Elements, this is a group action. Now, let $\mathbb S$ have $r$ orbits under this action. From Set of Orbits forms Partition, the orbits partition $\mathbb S$. Let these orbits be represented by $\set {S_1, S_2, \ldots, S_r}$, so that: {{begin-eqn}} {{eqn | l = \mathbb S       | r = \Orb {S_1} \cup \Orb {S_2} \cup \ldots \cup \Orb {S_r}       | c =  }} {{eqn | l = \size {\mathbb S}       | r = \size {\Orb {S_1} } + \card {\Orb {S_2} } + \ldots + \size {\Orb {S_r} }       | c =  }} {{end-eqn}} If each orbit had length divisible by $p$, then $p \divides N$. But this can not be the case, because, as we have seen: :$N \equiv k \pmod p$ So at least one orbit has length which is ''not'' divisible by $p$. Let $S \in \set {S_1, S_2, \ldots, S_r}$ be such that $\size {\Orb S)} = m: p \nmid m$. Let $s \in S$. It follows from Group Action on Prime Power Order Subset that: :$\Stab S s = S$ and so: :$\size {\Stab S} = \size S = p^n$ From Stabilizer is Subgroup: :$\Stab S \le G$ Thus $\Stab S$ is the subgroup of $G$ with $p^n$ elements of which we wanted to prove the existence. {{Qed}}
Let $\Bbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements. Now let $G$ act on $\Bbb S$ by the rule: :$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$ From Set of Orbits forms Partition, the orbits partition $\mathbb S$. Let these orbits be represented by $\set {S_1, S_2, \ldots, S_n}$, so that: :$(1): \quad \size {\mathbb S} = \size {\Orb {S_1} } + \card {\Orb {S_2} } + \ldots + \size {\Orb {S_r} }$ Thus each $\Orb {S_i}$ is the orbit under $*$ of some $S_i$ whose order is $p^n$. By the Orbit-Stabilizer Theorem: :$(2): \quad \order {\Orb {S_i} } = \dfrac {\order G} {\order {\Stab {S_i} } }$ for all $i \in \set {1, 2, \ldots, n}$ where $\Stab {S_i}$ is the stabilizer of $S_i$ under $*$. Let $s \in S_i$ and $x \in \Stab {S_i}$. Then $s x \in S_i$ because $S_i x = S_i$. Let $s$ be fixed while $x$ runs over $\Stab {S_i}$. Then: :$\forall s \in S_i: s \Stab {S_i} \subseteq S_i$ Therefore $S_i$ is the union of all the left cosets $s \Stab {S_i}$. These left cosets may not all be distinct. Let $r_i$ be the number of distinct left cosets of $s \Stab {S_i}$. Then: :$\card {S_i} = r_i \card {\Stab {S_i} }$ because: : distinct left cosets are disjoint : all have the same cardinality $\card {\Stab {S_i} }$. Thus: :$(3): \quad \card {\Stab {S_i} } \divides \card {S_i} = p^n$ Because the only divisors of $p^n$ are smaller powers of $p$: :$\card {\Stab {S_i} } = p^{n_i}$ for some $n_i \le n$, for $i = 1, 2, \ldots, r$. We can write: :$g = \order G = k p^n$ where $p$ does not divide $k$. From $(2)$ and $(3)$: :$\card {S_i} = k p^{d_i}$ where $d_i = n - n_i$ for $i = 1, 2, \ldots, r$. From Cardinality of Set of Subsets: :$\card {\Bbb S} = \dbinom {k p^n} {p_n}$ We have from Binomial Coefficient involving Power of Prime that: :$\dbinom {k p^n} {p_n} \equiv k \pmod p$ But $p$ does not divide $k$. Hence $p$ does not divide $\card {\Bbb S}$. Hence by $(1)$ and $(4)$: :$p \nmid k \paren {p^{d_1} + p^{d_2} + \cdots p^{d_r} }$ If $d_i > 0$ for all $i = 1, 2, \ldots, r$, then $p$ would divide this integer. So there must be some $i$ such that $d_i = 0$. Then, for this $i$: :$n_i = n$ That is: :$\card {\Stab {S_i} } = p^n$ Thus $\Stab {S_i}$ is the subgroup of $G$ of order $p_n$ that was asserted. {{qed}}
From Composition Series of Group of Prime Power Order, a $p$-group has subgroups corresponding to every divisor of its order. Thus, taken with the First Sylow Theorem, a finite group has a subgroup corresponding to every prime power divisor of its order. {{qed}}
Let $Q$ be a $p$-subgroup of $N = \map {N_G} P$. Let $\order Q = p^m, \order P = p^n$. By Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup: :$P \lhd \map {N_G} P$ thus by Subset Product with Normal Subgroup as Generator: :$\gen {P, Q} = P Q$ Thus by Order of Subgroup Product: :$P Q \le G: \order {P Q} = p^{n + m - s}$ where $\order {P \cap Q} = p^s$. Since $n$ is the highest power of $p$ dividing $\order G$, this is possible only when $m \le s$. Since $P \cap Q \le Q, s \le m$ thus we conclude that $m = s$ and therefore $P \cap Q = Q$. Thus from Intersection with Subset is Subset: :$Q \subseteq P$ In particular, if $Q$ is a Sylow $p$-subgroup of $\map {N_G} P$, then $Q = P$. {{qed}}
Let $P$ be a Sylow $p$-subgroup of $G$. Let $\mathbb S$ be the set of all distinct $G$-conjugates of $P$: :$\mathbb S = \set {g P g^{-1}: g \in G}$ Let $h * S$ be the conjugacy action: :$\forall h \in P, S \in \mathbb S: h * S = h S h^{-1}$ From Conjugacy Action on Subgroups is Group Action, this is a group action for $S \le G$. To show it is closed for $S \in \mathbb S$: {{begin-eqn}} {{eqn | l = S       | o = \in       | r = \mathbb S       | c =  }} {{eqn | ll= \leadsto       | l = \exists g \in G: S       | r = g P g^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = h * S       | r = h \paren {g P g^{-1} } h^{-1}       | c =  }} {{eqn | r = \paren {h g} P \paren {h g}^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = h * S       | o = \in       | r = \mathbb S       | c =  }} {{end-eqn}} So, consider the orbits and stabilizers of $\mathbb S$ under this group action. Since $\forall S \in \mathbb S: \Stab S \le P$, we have that: :$\size {\Stab S} \divides \order P$ Therefore, by the Orbit-Stabilizer Theorem, these orbit lengths are all congruent to either $0$ or $1$ modulo $p$, since $P$ is a Sylow $p$-subgroup of $G$. Note that this will imply, as we shall mark later on: :$\size {\mathbb S} \equiv 1 \pmod p$ Now, $h * P = h P h^{-1} = P$, so: :$\Orb P = \set P$ We now show that $P$ is the only element of $\mathbb S$ such that $\size {\Orb S} = 1$. If $g P g^{-1}$ has one element in its orbit, then: :$\forall x \in P: x \paren {g P g^{-1} } x^{-1} = g P g^{-1}$ Thus $\forall x \in P$ we have that: :$g^{-1} x g \in \map {N_G} P$ From Order of Conjugate Element equals Order of Element, we have that: :$\order {g^{-1} x g} = \order x$ Thus $P_1 = g^{-1} P g$ is a $p$-subgroup of $\map {N_G} P$. As $P$ and $P_1$ have the same number of elements, $P_1$ is a Sylow $p$-subgroup of $\map {N_G} P$. Hence $P_1 = P$ by Normalizer of Sylow p-Subgroup, so $g P g^{-1} = P$. Thus $P$ is the only element of $\mathbb S$ whose orbit has length $1$. From Stabilizer of Coset Action on Set of Subgroups, $P = \map {N_G} P$. Thus, for any $g \notin P$, $\size {\Orb {g P g^{-1} } }$ under conjugation by elements of $P$ has orbit greater than $1$. Hence: : $\size {\mathbb S} \equiv 1 \pmod p$ as promised. Next we consider orbits of $\mathbb S$ under conjugation by elements of $Q$. Since every orbit has length a power of $p$, the above conclusion shows there is at least one orbit of length $1$. So there is an element $g$ such that: :$\forall x \in Q: x \paren {g P g^{-1} } x^{-1} = g P g^{-1}$ As previously: :$g^{-1} Q g \subseteq \map {N_G} P$ So by Normalizer of Sylow p-Subgroup: :$g^{-1} Q g \subseteq P$ Thus $Q \subseteq g P g^{-1}$ as required. {{qed}}
Suppose $P$ and $Q$ are Sylow $p$-subgroups of $G$. By the Second Sylow Theorem, $Q$ is a subset of a conjugate of $P$. But since $\order P = \order Q$, it follows that $Q$ must equal a conjugate of $P$. {{qed}}