hcju/proofwikiqa huggingface dataset:
Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$. Let $H$ be a Sylow $p$-subgroup of $G$. We have that: :$\order H = p^n$ :$\index G H = m$ Let $S_1, S_2, \ldots, S_m$ denote the left cosets of $G \pmod H$. We have that $G$ acts on $G / H$ by the rule: :$g * S_i = g S_i$. Let $H_i$ denote the stabilizer of $S_i$. By the Orbit-Stabilizer Theorem: :$\order {H_i} = p^n$ while: :$S_i = g H \implies g H g^{-1} \subseteq H_i$ Because $\order {g H g^{-1} } = \order H = \order {H_i}$, we have: :$g H g^{-1} \subseteq H_i$ Let $H'$ be a second Sylow $p$-subgroup of $G$. Then $H'$ acts on $G / H$ by the same rule as $G$. Since $p \nmid m$, there exists at least one orbit under $H'$ whose cardinality is not divisible by $p$. Suppose that $S_1, S_2, \ldots, S_r$ are the elements of an orbit where $p \nmid r$. Let $K = H' \cap H_1$. Then $K$ is the stabilizer of $S_1$ under the action of $H'$. Therefore: :$\index {H'} K = r$ However: :$\order {H'} = p^n$ and: :$p \nmid r$ from which it follows that: :$r = 1$ and: :$K = H'$ Therefore: :$\order K = \order {H'} = \order {H_1} = p^n$ and: :$H' = K = H_1$ Thus $H'$ and $H$ are conjugates.
Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$. Let $r$ be the number of Sylow $p$-subgroups of $G$. Let $H$ be a Sylow $p$-subgroup of $G$. We have that: :$\order H = p^n$ :$\index G H = m$ Let $S_1, S_2, \ldots, S_m$ denote the elements of the left coset space of $G / H$. We have that $H$ acts on $G / H$ by the rule: :$g * S_i = g S_i$ for $S_i \in G / H$. Unless $H = G$ and $r = 1$, there is more than $1$ orbit. {{explain|Why? See Orbits of Group Action on Sets with Power of Prime Size}} We have that $H$ is the stabilizer of the coset $H$, which must be one of $S_1, S_2, \ldots, S_m$. Let $S_1, S_2, \ldots, S_k$ be the elements of $G / H$ whose stabilizer is $H$. From the Orbit-Stabilizer Theorem and from $\order H = p^n$ we see there are $2$ cases: :$(1): \quad$ The orbit of $S_i$ contains $p^t$ elements where $0 < t < n$ :$(2): \quad$ The orbit of $S_i$ contains only the element $S_i$. $(2)$ occurs {{iff}} $S_i$ is one of the cosets $S_1, S_2, \ldots, S_k$ whose stabilizer is $H$. So counting the elements of $G / H$, we see that: :$m = k + u p$ or: :$m \equiv k \pmod p$ From the Fifth Sylow Theorem, we have: :$m \equiv k r \pmod p$ and so: :$k r \equiv k \pmod p$ from which it follows: :$r \equiv 1 \pmod p$ because $k \not \equiv 0 \pmod p$. Hence the result. {{qed}}
If $G$ has precisely one Sylow $p$-subgroup, it must be normal from Unique Subgroup of a Given Order is Normal. Suppose a Sylow $p$-subgroup $P$ is normal. Then it equals its conjugates. Thus, by the Third Sylow Theorem, there can be only one such Sylow $p$-subgroup. {{qed}}
Since $N \lhd G$, we see that: :$\gen {P, N} = P N$ from Subset Product with Normal Subgroup as Generator. Since $P \cap N \le P$, it follows that: :$\order {P \cap N} = p^k$ where $k > 0$. By Order of Subgroup Product: :$\order {P N} \order {P \cap N} = \order P \order N$ Hence from Lagrange's Theorem: :$\index N {P \cap N} = \index {P N} P$ By Tower Law for Subgroups: :$\index G P = \index G {P N} \index {P N} P$ Thus: :$\index {P N} P \divides \index G P$ where $\divides$ denotes divisibility. Thus: :$p \nmid \index {P N} P$ so: :$p \nmid \index N {P \cap N}$ Thus $P \cap N$ is a Sylow $p$-subgroup of $N$. {{qed}}
We have that $P \le G$ and $ N \lhd G$. So by the Second Isomorphism Theorem for Groups: :$P N / N \cong P / \paren {P \cap N}$ We have that: :$P N / N = \set {p N : p \in P}$ and so every element of $P N / N$ has order a power of $p$. Hence $P N / N$ is a $p$-subgroup of $G / N$. From Intersection of Normal Subgroup with Sylow P-Subgroup, we have that: :$p \nmid \index G {P N}$ So $P N / N$ is a Sylow $p$-subgroup of $G / N$. {{qed}}
Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$. Let $r$ be the number of Sylow $p$-subgroups of $G$. It is to be shown that $r \divides m$. Let $H$ be a Sylow $p$-subgroup of $G$. We have that: :$\order H = p^n$ :$\index G H = m$ Let $S_1, S_2, \ldots, S_m$ denote the elements of the left coset space of $G / H$. We have that $G$ acts on $G / H$ by the rule: :$g * S_i = g S_i$ for $S_i \in G / H$. There is only one orbit under this action, namely the whole of $G / H$. {{explain|Why? See Orbits of Group Action on Sets with Power of Prime Size}} Therefore the stabilizer of each $S_i$ is a subgroup of $G$ of index $m$ and order $p^n$. In other words, each $S_i$ has a Sylow $p$-subgroup as a stabilizer. Now it is shown that each Sylow $p$-subgroup is the stabilizer of one or more of the cosets $S_1, S_2, \ldots, S_m$. We have that $H$ is the stabilizer of the coset $H$, which must be one of $S_1, S_2, \ldots, S_m$. Let $S_1, S_2, \ldots, S_k$ be the elements of $G / H$ whose stabilizer is $H$. By the Third Sylow Theorem, any other Sylow $p$-subgroup of $H$ is a conjugates $g H g^{-1}$ of $H$. Thus it is seen that $g H g^{-1}$ is a stabilizer of the cosets $g S_1, g S_2, \ldots, g S_k$. So each of the $r$ distinct Sylow $p$-subgroups of $G$ is the stabilizer of exactly $k$ elements of $G / H$. Thus: :$m = k r$ and so: :$r \divides m$ as required. {{qed}}
Let $\left({G, +, \circ}\right)_R$ be an $R$-module and $S$ be a set. We need to show that $\left({G^{\left({S}\right)}, +'}\right)$ is a group. Let $f, g \in G^{\left({S}\right)}$. Let: : $F = \left\{{x \in S: f \left({x}\right) \ne e}\right\}$ : $G = \left\{{x \in S: g \left({x}\right) \ne e}\right\}$ From the definition of $f$ and $g$, both $F$ and $G$ are finite. Since $e + e = e$ by definition of identity element, it follows that if: : $f \left({x}\right) + g \left({x}\right) \ne e$ then necessarily $f \left({x}\right) \ne e$ or $g \left({x}\right) \ne e$. That is: : $\left({f +' g}\right) \left({x}\right) = f \left({x}\right) + g \left({x}\right) \ne e \implies x \in F \cup G$ But as $F$ and $G$ are both finite, it follows that $F \cup G$ is also finite. Hence $f +' g \in G^{\left({S}\right)}$ and $\left({G^{\left({S}\right)}, +'}\right)$ is closed. Now let $f \in G^{\left({S}\right)}$. Let $f^*$ be the Induced Structure Inverse of $f$. Thus $f^* \left({x}\right) = - \left({f \left({x}\right)}\right)$. Again, let $F = \left\{{x \in S: f \left({x}\right) \ne e}\right\}$. It follows directly that $x \in S \setminus F \implies f^* \left({x}\right) = e$. Hence $f^* \left({x}\right) \ne e \implies x \in F$ and hence $f^* \in G^{\left({S}\right)}$. So by the Two-Step Subgroup Test, it follows that $\left({G^{\left({S}\right)}, +'}\right)$ is a subgroup of $\left({G^S, +}\right)$. Hence $\left({G^{\left({S}\right)}, +', \circ}\right)_R$ is an $R$-module. The result follows. {{qed}}
Since a linear transformation $\phi: G \to H$ is, in particular, a homomorphism from the group $G$ to the group $H$, it follows that: :$(1): \quad$ By Homomorphism with Cancellable Codomain Preserves Identity, $\map \phi {e_G} = e_H$ :$(2): \quad$ By Homomorphism with Identity Preserves Inverses, $\map \phi {-x} = -\map \phi x$. From Epimorphism preserves Modules and definition of surjection, it follows that as $M$ is a submodule of $G$, then $\phi \sqbrk M$ is a submodule of $H$. The result follows ... {{stub}}
Let $x, y \in S$. Let $\lambda \in R$. Then: {{begin-eqn}} {{eqn | l = \map \phi {x + y}       | r = \map \phi x + \map \phi y       | c = {{Defof|Linear Transformation}} }} {{eqn | r = \map \psi x + \map \psi y       | c = $x, y \in S$ }} {{eqn | r = \map \psi {x + y}       | c = {{Defof|Linear Transformation}} }} {{eqn | l = \map \phi {\lambda \circ x}       | r = \lambda \circ \map \phi x       | c = {{Defof|Linear Transformation}} }} {{eqn | r = \lambda \circ \map \psi x       | c = $x \in S$ }} {{eqn | r = \map \psi {\lambda \circ x}       | c = {{Defof|Linear Transformation}} }} {{end-eqn}} Hence $x + y, \lambda \circ x \in S$. By Submodule Test, $S$ is a submodule of $G$. {{qed}}
The proof proceeds on several dimensional-cases. Note that the case $m = 3$ is incredibly intricate, and that a full proof would be impractical to produce here.  An outline of the $m = 3$ case will be given instead. === Dimension $m = 1$ === Follows from the Classification of Compact One-Manifolds. {{qed|lemma}} === Dimension $m = 2$ === Follows from the Classification of Compact Two-Manifolds. {{qed|lemma}} === Dimension $m = 3$ === Follows from Thurston's Geometrization Conjecture, proved by {{AuthorRef|Grigori Perelman}}. {{ProofWanted}} {{qed|lemma}} === Dimension $m = 4$ === Follows from $4$-dimensional Topological h-Cobordism Theorem of {{AuthorRef|Andrew John Casson}} and {{AuthorRef|Michael Hartley Freedman}}. {{ProofWanted}} {{qed|lemma}} === Dimension $m = 5$ === {{ProofWanted}} Summary: Any $\Sigma^5$ bounds a contractible $6$-manifold $Z$. Let $\Bbb D^6$ be a $6$-disk (AKA 6-ball). {{explain|Open or closed disk / ball?}} Then $Z - \Bbb D^6$ is an $h$-cobordism between $\Sigma$ and $\partial \Bbb D^6 = \Bbb S^5$. Hence $\Sigma$ is differomorphic to $\Bbb S^5$ by the h-Cobordism Theorem. {{qed|lemma}} === Dimension $m \ge 6$ === {{ProofWanted}} We can cut two small $m$-disks $D', D''$ from $\Sigma$. The remaining manifold, $\Sigma \setminus \paren {D' \cup D''}$ is an h-cobordism between $\partial D'$ and $\partial D''$. These are just two copies of $\Bbb S^{m-1}$. By the $h$-cobordism theorem, there exists a diffeomorphism: :$\phi: \Sigma \setminus \paren {D' \cup D''} \to \Bbb S^{m - 1} \times \closedint 0 1$ which can be chosen to restrict to the identity on one of the $\Bbb S^{m - 1}$. Let $\Xi$ denote this $\Bbb S^{m - 1}$ such that $\phi$ restricts to the identity. Since $\psi \vert_\Xi = Id$, we can extend $\psi$ across $D''$, the interior of $\Xi$ to obtain a diffeomorphism $\phi': \Sigma \setminus D'' \to \Bbb S^{m - 1} \cup D'$. {{explain|What does $\psi \vert_\Xi$ mean?}} Let $\Bbb D^m$ denote this latter manifold, which is merely an $m$-disk. Our diffeomorphism $\phi': \Sigma \setminus D'' \to \Bbb D^m$ induces a diffeomorphism on the boundary spheres $\Bbb S^{m - 1}$. Any diffeomorphism of the boundary sphere $\Bbb S^{m - 1}$ can be extended radially to the whole disk: :$\map {\operatorname {int} } {\Bbb S^{m - 1} } = D''$ but only as a homeomorphism of D''.   Hence the extended function $\phi'': \Sigma \to \Bbb S^m$ is a homeomorphism. {{qed}} {{Namedfor|Jules Henri Poincaré|cat = Poincaré}}
First the extreme case: The smallest submodule of $G$ containing $\O$ is $\set {e_G}$. By definition, $\set {e_G}$ is the set of all linear combinations of $\varnothing$. Now the general case: Let $\O \subset S \subseteq G$. Let $L$ be the set of all linear combinations of $S$. Since $G$ is a unitary $R$-module, every element $x \in S$ is the linear combination $1_R x$, so $S \subseteq L$. But $L$ is closed for addition and scalar multiplication, so is a submodule. Thus $H \subseteq L$. But as every linear combination of $S$ clearly belongs to any submodule of $G$ which contains $S$, we also have $L \subseteq H$.
There are no sequences at all of $n$ terms of the empty set for any $n > 0$. Hence the result holds vacuously. {{qed}}
From Scalar Product with Identity, $\forall \lambda: \lambda \circ e = e$. Let $H \subseteq G$ such that $e \in H$. Consider any sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ in $H$ which includes $e$. So, let $a_j = e$ for some $j \in \closedint 1 n$. Let $c \in R \ne 0_R$. Consider the sequence $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ of elements of $R$ defined as: :$\lambda_k = \begin{cases} c & : k \ne j \\ 0_R & : k= j \end{cases}$ Then: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \lambda_k \circ a_k       | r = \lambda_1 \circ a_1 + \lambda_2 \circ a_2 + \cdots + \lambda_j \circ a_j + \cdots + \lambda_n \circ a_n       | c =  }} {{eqn | r = 0_R \circ a_1 + 0_R \circ a_2 + \cdots + c \circ e +  \cdots + 0_R \circ a_n       | c =  }} {{eqn | r = e + e + \cdots + e + \cdots + e       | c =  }} {{eqn | r = e       | c =  }} {{end-eqn}} Thus there exists a sequence $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ in which not all $\lambda_k = 0_R$ such that: :$\displaystyle \sum_{k \mathop = 1}^n \lambda_k \circ a_k = e$ Hence the result. {{qed}}
Let $G$ be an unitary $R$-module. Then $\sequence {a_n}$ is a linearly independent sequence {{iff}} $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$. So suppose that $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$. Then clearly $\sequence {a_n}$ is a linearly independent sequence of $G$. Conversely, let $\sequence {a_n}$ be a linearly independent sequence of $G$. Let $\sequence {b_m}$ be a sequence of distinct terms of $\set {a_1, a_2, \ldots, a_n}$. Let $\sequence {\mu_m}$ be a sequence of scalars such that $\displaystyle \sum_{j \mathop = 1}^m \mu_j b_j = 0$. For each $k \in \closedint 1 n$, let: :$\lambda_k = \begin{cases} \mu_j & : j \text { is the unique index such that } a_k = b_j \\ 0 & : a_k \notin \set {b_1, b_2, \ldots, b_m} \end{cases}$ Then: :$\displaystyle 0 = \sum_{j \mathop = 1}^m \mu_j b_j = \sum_{k \mathop = 1}^n \lambda_k a_k$ Thus: :$\forall k \in \closedint 1 n: \lambda_k = 0$ As $\set {\mu_1, \ldots, \mu_m} \subseteq \set {\lambda_1, \ldots, \lambda_n}$, it follows that: :$\forall j \in \closedint 1 m: \mu_j = 0$ and so $\sequence {b_m}$ has been shown to be a linearly independent sequence. Hence the result. {{Qed}}
Suppose $\left\{{a_1, a_2, \ldots, a_n}\right\}$ is a linearly independent set. Then by Subset of Linearly Independent Set, any subset $\left\{{b_1, b_2, \ldots, b_m}\right\}$ of $\left\{{a_1, a_2, \ldots, a_n}\right\}$ must itself be linearly independent. Thus if $\left\{{b_1, b_2, \ldots, b_m}\right\}$ is linearly dependent, then so must $\left\{{a_1, a_2, \ldots, a_n}\right\}$ be. {{qed}}
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of distinct terms of $A$. Let $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ a sequence of scalars. Then: $\displaystyle \sum_{k \mathop = 1}^n \lambda_k f_{a_k}$ is the mapping whose value at $a_k$ is $\lambda_k$ and whose value at any $x$ not in $\set {a_1, a_2, \ldots, a_n}$ is zero. Hence $B$ is a generator of $R^{\left({A}\right)}$ which is linearly independent. Thus, by definition, $B$ is a basis of $R^{\paren A}$. If $A = \closedint 1 n$, then $B$ is the standard basis of $R^n$. {{Qed}}
=== Necessary Condition === Let $\sequence {a_n}$ be an ordered basis of $G$. Then every element of $G$ is a linear combination of $\set {a_1, \ldots, a_n}$, which is a generator of $G$, by Generated Submodule is Linear Combinations. Thus there exists at least one such sequence of scalar. Now suppose there were two such sequences of scalars: $\sequence {\lambda_n}$ and $\sequence {\mu_n}$. That is, suppose $\displaystyle \sum_{k \mathop = 1}^n \lambda_k a_k = \sum_{k \mathop = 1}^n \mu_k a_k$. Then: {{begin-eqn}} {{eqn | l =\ sum_{k \mathop = 1}^n \paren {\lambda_k - \mu_k} a_k       | r = \sum_{k \mathop = 1}^n \paren {\lambda_k a_k - \mu_k a_k}       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n \lambda_k a_k - \sum_{k \mathop = 1}^n \mu_k a_k       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So $\lambda_k = \mu_k$ for all $k \in \closedint 1 n$ as $\sequence {a_n}$ is a linearly independent sequence. {{qed|lemma}} === Sufficient Condition === Now suppose there is one and only one sequence $\sequence {\lambda_n}$ such that the condition holds. It is clear that $\set {a_1, \ldots, a_n}$ generates $G$. Suppose $\displaystyle \sum_{k \mathop = 1}^n \lambda_k a_k = 0$. Then, since also $\displaystyle \sum_{k \mathop = 1}^n 0 a_k = 0$, we have, by hypothesis: :$\forall k \in \closedint 1 n: \lambda_k = 0$ Therefore $\sequence {a_n}$ is a linearly independent sequence. {{qed}}
By Unique Representation by Ordered Basis, $\psi$ is a bijection. We have: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \lambda_k a_k + \sum_{k \mathop = 1}^n \mu_k a_k       | r = \sum_{k \mathop = 1}^n \paren {\lambda_k a_k + \mu_k a_k}       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n \paren {\lambda_k + \mu_k} a_k       | c =  }} {{end-eqn}} and we have: {{begin-eqn}} {{eqn | l = \beta \sum_{k \mathop = 1}^n \lambda_k a_k       | r = \sum_{k \mathop = 1}^n \beta \paren {\lambda_k a_k}       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n \paren {\beta \lambda_k} a_k       | c =  }} {{end-eqn}} thus proving that $\psi$ is also a homomorphism. {{Qed}}
From Isomorphism from R^n via n-Term Sequence, they are both isomorphic to the $R$-module $R^n$. {{qed}}
Follows directly from Standard Ordered Basis is Basis. {{qed}} Category:Module on Cartesian Product 4e2u0vlwjpxp8kx7zj4kipknso5ie65
By Isomorphism from R^n via n-Term Sequence, the mapping $\phi: G \to H$ defined as: : $\displaystyle \phi \left({\sum_{k \mathop = 1}^n \lambda_k a_k}\right) = \sum_{k \mathop = 1}^n \lambda_k b_k$ is well-defined. Thus: : $\forall k \in \left[{1 \,.\,.\, n}\right]: \phi \left({a_k}\right) = b_k$ {{wtd|Verification that $\phi$ is linear needs to be done.}} By Linear Transformation of Generated Module, $\phi$ is the only linear transformation whose value at $a_k$ is $b_k$ for all $k \in \left[{1 \,.\,.\, n}\right]$. {{Qed}}
Suppose: :$\displaystyle \sum_{k \mathop = 1}^r \lambda_k \phi \left({a_k}\right) = 0$ Then: :$\displaystyle \phi \left({\sum_{k \mathop = 1}^r \lambda_k a_k}\right) = 0$ So $\displaystyle \sum_{k \mathop = 1}^r \lambda_k \phi \left({a_k}\right)$ belongs to the kernel of $\phi$ and hence is also a linear combination of $\left\{{a_k: r + 1 \le k \le n}\right\}$. Thus $\forall k \in \left[{1 \,.\,.\, r}\right]: \lambda_k = 0$ since $\left \langle {a_n} \right \rangle$ is linearly independent. Thus the sequence $\left \langle {\phi \left({a_r}\right)} \right \rangle$ is linearly independent. We have $\forall k \in \left[{r + 1 \,.\,.\, n}\right]: \phi \left({a_k}\right) = 0$. So let $x \in G$. Let $\displaystyle x = \sum_{k \mathop = 1}^n \mu_k a_k$. Then: : $\displaystyle \phi \left({x}\right) = \sum_{k \mathop = 1}^n \mu_k \phi \left({a_k}\right) = \sum_{k \mathop = 1}^r \mu_k \phi \left({a_k}\right)$ Therefore $\left \langle {\phi \left({a_r}\right)} \right \rangle$ is an ordered basis of the image of $\phi \left({G}\right)$. {{Qed}}
From the definition of a module, the group $\left({H, +_H}\right)$ is abelian. Therefore we can apply Inverse Mapping in Induced Structure to show that $- \phi: G \to H$ is a homomorphism. Then: {{begin-eqn}} {{eqn | l=\left({- \phi}\right) \left({\lambda \circ x}\right)       | r=-\phi \left({\lambda \circ x}\right)       | c= }} {{eqn | r=- \lambda \circ \phi \left({x}\right)       | c= }} {{eqn | r=\lambda \circ \left({- \phi \left({x}\right)}\right)       | c= }} {{eqn | r=\lambda \circ \left({- \phi}\right) \left({x}\right)       | c= }} {{end-eqn}} {{qed}}
From Structure Induced by Group Operation is Group, $\struct {H^G, \oplus_H}$ is a group Let $\phi, \psi \in \map {\LL_R} {G, H}$. From Addition of Linear Transformations, $\phi \oplus_H \psi \in \map {\LL_R} {G, H}$. From Negative Linear Transformation, $-\phi \in \map {\LL_R} {G, H}$. Thus, from the Two-Step Subgroup Test, $\struct {\map {\LL_R} {G, H}, \oplus_H}$ is a subgroup of $\struct {H^G, \oplus_H}$. {{finish|Still need to show that it is abelian}}
Let $\lambda \in \lambda \in \map Z R$. Then: {{begin-eqn}} {{eqn | l = \map {\paren {\lambda \circ \phi} } {x +_G y}       | r = \lambda \circ \map \phi {x +_G y}       | c =  }} {{eqn | r = \lambda \circ \paren {\map \phi x +_H \map \phi y}       | c =  }} {{eqn | r = \lambda \circ \map \phi x +_H \lambda \circ \map \phi y       | c =  }} {{eqn | r = \map {\paren {\lambda \circ \phi} } x +_H \map {\paren {\lambda \circ \phi} } y       | c =  }} {{end-eqn}} Because $\lambda \in \map Z R$, $\lambda$ commutes with all elements of $R$. So $\forall \mu \in R: \lambda \circ \mu = \mu \circ \lambda$. Thus: {{begin-eqn}} {{eqn | l = \map {\paren {\lambda \circ \phi} } {\mu \circ x}       | r = \lambda \circ \map \phi {\mu \circ x}       | c =  }} {{eqn | r = \lambda \circ \mu \circ \map \phi x       | c =  }} {{eqn | r = \mu \circ \lambda \circ \map \phi x       | c = as $\lambda \in \map Z R$ }} {{eqn | r = \mu \circ \map {\paren {\lambda \circ \phi} } x       | c =  }} {{end-eqn}} {{qed}}
From Group equals Center iff Abelian, the center of a commutative ring is the entire ring. The result follows from Linear Transformation from Center of Scalar Ring. {{qed}}
Let $B = \set {\phi_{i j}: i \in \closedint 1 n, j \in \closedint 1 m}$. Let $\displaystyle \sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \lambda_{i j} \phi_{i j} = 0$. Then: :$\displaystyle \forall k \in \closedint 1 n: 0 = \sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \lambda_{i j} \map {\phi_{i j} } {a_k} = \sum_{j \mathop = 1}^m \lambda_{k j} b_j$ So: :$\forall j \in \closedint 1 n: \lambda_{k j} = 0$ Hence $B$ is linearly independent. Now let $\phi \in \map {\LL_R} {G, H}$. Let $\tuple {\alpha_{i 1}, \alpha_{i 2}, \ldots, \alpha_{i m} }$ be the sequence of scalars that satisfies: :$\displaystyle \forall i \in \closedint 1 n: \map \phi {a_i} = \sum_{j \mathop = 1}^m \alpha_{i j} b_j$ Then: :$\displaystyle \forall k \in \closedint 1 n: \map \phi {a_k} = \map {\paren {\sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \alpha_{i j} u_{i j} } } {a_k}$ by a calculation similar to the preceding. So, by Linear Transformation of Generated Module: :$\displaystyle u = \sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \alpha_{i j} u_{i j}$ Thus $B$ is a generator for $\phi \in \map {\LL_R} {G, H}$. {{Qed}}
Follows from Composite of R-Algebraic Structure Homomorphisms is Homomorphism, as it is a subring of the ring of all endomorphisms of the abelian group $\struct {G, +}$. {{ProofWanted}} Category:Linear Operators eda5k9ojbrvu9i00azjlutffjtyrjib
As $G$ is finite-dimensional, then by Evaluation Isomorphism is Isomorphism $J: G \to G^{**}$ is an isomorphism, and therefore a surjection. Thus: :$N^\circ = \left\{{x^\wedge \in G^{**}: \forall t' \in N: x^\wedge \left({t'}\right) = 0}\right\}$ where $x^\wedge$ is as defined in the definition of the evaluation linear transformation. The result follows. {{qed}}
A vector space is a module, so all results about modules also apply to vector spaces. So from Scalar Product with Identity it follows directly that: :$\lambda = 0_F \lor \mathbf v = e \implies \lambda \circ \mathbf v = \bszero$ Next, suppose $\lambda \circ \mathbf v = \bszero$ but $\lambda \ne 0_F$. Then: {{begin-eqn}} {{eqn | l = \bszero       | r = \lambda^{-1} \circ \bszero       | c = Zero Vector Scaled is Zero Vector }} {{eqn | r = \lambda^{-1} \circ \paren {\lambda \circ \mathbf v}       | c = as $\lambda \circ \mathbf v = \bszero$ }} {{eqn | r = \paren {\lambda^{-1} \circ \lambda} \circ \mathbf v       | c = {{Vector-space-axiom|7}} }} {{eqn | r = 1 \circ \mathbf v       | c = {{Field-axiom|M4}} }} {{eqn | r = \mathbf v       | c = {{Vector-space-axiom|8}} }} {{end-eqn}} {{Qed}}
The sufficient condition is proved in Vector Scaled by Zero is Zero Vector, and in Zero Vector Scaled is Zero Vector. The necessary condition is proved in Vector Product is Zero only if Factor is Zero. {{qed}}
Let us write $\phi \sqbrk V$ for the homomorphic image of $\phi$. From Homomorphic Image of R-Module is R-Module, $\phi \sqbrk V$ is a $K$-module. It thus suffices to show that $\phi \sqbrk V$ is unitary, since then it will be a $K$-vector space. To this end, let $1_K$ be the unity of $K$. Then for any $\map \phi {\mathbf v} \in \phi \sqbrk V$, compute: {{begin-eqn}} {{eqn | l = 1_K \circ_W \map \phi {\mathbf v}       | r = \map \phi {1_K \circ_V \mathbf v}       | c = $\phi$ is a linear transformation }} {{eqn | r = \map \phi {\mathbf v}       | c = $V$ is a $K$-vector space }} {{end-eqn}} Hence $\phi \sqbrk V$ is unitary, and so a $K$-vector space. {{qed}}