hcju/proofwikiqa huggingface dataset:
As $\struct {S, \circ}$ is a semigroup, it is already closed and associative. It remains to be shown that it has an identity and that every element of $S$ has an inverse in $S$. First we show that $\struct {S, \circ}$ has an identity. Choose $a \in S$. Let the mapping $\lambda_a: S \to S$ be the left regular representation of $\struct {S, \circ}$ with respect to $a$. By hypothesis: :all elements of $S$ are cancellable :$S$ is finite. By Regular Representation wrt Cancellable Element on Finite Semigroup is Bijection, $\lambda_a$ is a bijection. Hence $a \circ e = a$ for some $e \in S$. Let $x \in S$. Then because of cancellability: {{begin-eqn}} {{eqn | l = a \circ e \circ x       | r = a \circ x       | c =  }} {{eqn | ll= \leadsto       | l = e \circ x       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = x \circ e \circ x       | r = x \circ x       | c =  }} {{eqn | ll= \leadsto       | l = x \circ e       | r = x       | c =  }} {{end-eqn}} Thus $e$ is the identity. The existence of inverses comes from the surjectivity of $\lambda_a$. As $\lambda_a$ is surjective: :$\exists y \in S: \map {\lambda_a} y = e$ That is: :$a \circ y = e$ So we see that $y$ acts as a right inverse for $a$. This is the case for any $a \in S$: all of them have some right inverse. So, from Right Inverse for All is Left Inverse, each of these elements is also a left inverse, and therefore an inverse. Thus $S$ is closed, associative, has an identity and every element has an inverse. So, by definition, $\struct {S, \circ}$ is a group. {{qed}}
List the positive powers $x, x^2, x^3, \ldots$ of any element $x$ of a finite semigroup $\left({S, \circ}\right)$. Since all are elements of $S$, and the semigroup has a finite number of elements, it follows from the Pigeonhole Principle this list must contain repetitions. So there must be at least one instance where $x^m = x^n$ for some $m, n \in \N$. {{Qed}} Category:Semigroups 1xgxx01rhazfn21f659gs2m7pt8z620
From Finite Semigroup Equal Elements for Different Powers, we have: :$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$ Let $m > n$. Let $n = k, m = k + l$. Then: : $\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$ Now we show that: :$x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$ That is, that $x^{k l}$ is idempotent. First: {{begin-eqn}} {{eqn | l = x^k       | r = x^{k + l}       | c =  }} {{eqn | ll= \leadsto       | l = x^k \circ x^l       | r = x^{k + l} \circ x^l       | c = both sides ${} \circ x^l$ }} {{eqn | ll= \leadsto       | l = x^{k + l}       | r = x^{k + 2 l}       | c = Index Laws for Semigroup: Sum of Indices }} {{eqn | ll= \leadsto       | l = x^k       | r = x^{k + 2 l}       | c = as $x^k = x^{k + l}$ }} {{end-eqn}} From here we can easily prove by induction that: : $\forall n \in \N: x^k = x^{k + n l}$ In particular: : $x^k = x^{k + k l} = x^{k \paren {l + 1} }$ There are two cases to consider: :$(1): \quad$ If $l = 1$, then $x^k = x^{k \paren {l + 1} } = x^{2 k} = x^k \circ x^k$, and $x^{k l} = x^k$ is idempotent. :$(2): \quad$ If $l > 1$, then: {{begin-eqn}} {{eqn | l = x^k       | r = x^{k + k l}       | c =  }} {{eqn | r = x^{k \paren {l + 1} }       | c =  }} {{eqn | ll= \leadsto       | l = x^k \circ x^{k \paren {l - 1} }       | r = x^{k \paren {l + 1} } \circ x^{k \paren {l - 1} }       | c = both sides ${} \circ x^{k \paren {l - 1} }$ }} {{eqn | ll= \leadsto       | l = x^{k + k l - k}       | r = x^{kl + k + k l - k}       | c = Index Laws for Semigroup: Sum of Indices }} {{eqn | ll= \leadsto       | l = x^{k l}       | r = x^{k l + k l}       | c =  }} {{eqn | r = x^{k l} \circ x^{k l}       | c =  }} {{end-eqn}} and again, $x^{k l}$ is idempotent. {{Qed}}
Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition: :$\paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$ First we verify $\map P 0$. When $n = 0$, we have: {{begin-eqn}} {{eqn | l = \paren {0 \cdot x} \circ x       | r = 0_R \circ x       | c =  }} {{eqn | r = 0_R       | c =  }} {{eqn | r = 0 \cdot \paren {x \circ x}       | c =  }} {{eqn | r = x \circ 0_R       | c =  }} {{eqn | r = x \circ \paren {0 \cdot x}       | c =  }} {{end-eqn}} So $\map P 0$ holds. === Basis for the Induction === Now we verify $\map P 1$: {{begin-eqn}} {{eqn | l = \paren {1 \cdot x} \circ x       | r = x \circ x       | c =  }} {{eqn | r = 1 \cdot \paren {x \circ x}       | c =  }} {{eqn | r = x \circ \paren {1 \cdot x}       | c =  }} {{end-eqn}} So $\map P 1$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\paren {k \cdot x} \circ x = k \cdot \paren {x \circ x} = x \circ \paren {k \cdot x}$ Then we need to show: :$\paren {\paren {k + 1} \cdot x} \circ x = \paren {k + 1} \cdot \paren {x \circ x} = x \circ \paren {\paren {k + 1} \cdot x}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {\paren {k + 1} \cdot x} \circ x       | r = \paren {x + \paren {k \cdot x} } \circ x       | c =  }} {{eqn | r = x \circ x + \paren {k \cdot x} \circ x       | c = {{Ring-axiom|D}} }} {{eqn | r = x \circ x + k \cdot \paren {x \circ x}       | c = Induction Hypothesis }} {{eqn | r = \paren {k + 1} \cdot \paren {x \circ x}       | c =  }} {{end-eqn}} A similar construction shows that $\paren {k + 1} \cdot \paren {x \circ x} = x \circ \paren {\paren {k + 1} \cdot x}$. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: \paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$ {{qed|lemma}} The result for $n < 0$ follows directly from Powers of Group Elements. {{qed}} Category:Ring Theory Category:Proofs by Induction 8hv2385mggjxed224pcgpzdu9xdduss
Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition $y^n = a^{-1} \circ x^n \circ a$. $\map P 0$ is true, as this just says $e =  a^{-1} \circ e \circ a$. === Basis for the Induction === $\map P 1$ is the case $y = a^{-1} \circ x \circ a$, which is how conjugacy is defined for $x$ and $y$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$y^k = a^{-1} \circ x^k \circ a$ Then we need to show: :$y^{k + 1} = a^{-1} \circ x^{k + 1} \circ a$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = y^{k + 1}       | r = \paren {a^{-1} \circ x \circ a}^{k + 1}       | c =  }} {{eqn | r = \paren {a^{-1} \circ x \circ a}^k \circ \paren {a^{-1} \circ x \circ a}       | c =  }} {{eqn | r = \paren {a^{-1} \circ x^k \circ a} \circ \paren {a^{-1} \circ x \circ a}       | c = Induction Hypothesis }} {{eqn | r = a^{-1} \circ x^k \circ x \circ a       | c =  }} {{eqn | r = a^{-1} \circ x^{k + 1} \circ a       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: y^n = a^{-1} \circ x^n \circ a$ Now we need to show that if $\map P n$ holds, then $\map P {-n}$ holds. That is: :$y^{-n} = a^{-1} \circ x^{-n} \circ a$ Let $n \in \N$. Then: {{begin-eqn}} {{eqn | l = y^{-n}       | r = \paren {a^{-1} \circ x \circ a}^{-n}       | c =  }} {{eqn | r = \paren {\paren {a^{-1} \circ x \circ a}^n}^{-1}       | c = Index Laws for Monoids: Negative Index }} {{eqn | r = \paren {a^{-1} \circ x^n \circ a}^{-1}       | c =  }} {{eqn | r = a^{-1} \circ \paren {x^n}^{-1} \circ \paren {a^{-1} }^{-1}       | c = Inverse of Group Product }} {{eqn | r = a^{-1} \circ x^{-n} \circ a       | c =  }} {{end-eqn}} Thus $\map P n$ has been shown to hold for all $n \in \Z$. Hence the result. {{qed}}
Proof by induction: For all $n \in \Z$, let $P \left({n}\right)$ be the proposition $a^n b = b a^{-n}$. $P(0)$ is trivially true, as $a^0 b = e b = b = b e = b a^{-0}$. === Basis for the Induction === $P(1)$ is true, as this is the given relation between $a$ and $b$: :$a b = b a^{-1}$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$a^k b = b a^{-k}$ Then we need to show: :$a^{k+1} b = b a^{-\left({k+1}\right)}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l=a^{k+1} b       | r=a \left({a^k b}\right)       | c= }} {{eqn | r=a \left({b a^{-k} }\right)       | c=Induction Hypothesis }} {{eqn | r=\left({a b}\right) a^{-k}       | c= }} {{eqn | r=\left({b a^{-1} }\right) a^{-k}       | c=Basis for the Induction }} {{eqn | r=b \left({a^{-1} a^{-k} }\right)       | c= }} {{eqn | r=b a^{-\left({k+1}\right)}       | c= }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore $\forall n \in \N: a^n b = b a^{-n}$. Now we show that $P \left({-1}\right)$ holds, i.e. that $a^{-1} b = b a$. {{begin-eqn}} {{eqn | l=a b       | r=b a^{-1}       | c= }} {{eqn | ll=\iff       | l=a b a       | r=b       | c= }} {{eqn | ll=\iff       | l=b a       | r=a^{-1} b       | c= }} {{end-eqn}} ... thus showing that $P \left({-1}\right)$ holds. The proof that $P \left({n}\right)$ holds for all $n \in \Z: n < 0$ then follows by induction, similarly to the proof for $n > 0$. {{qed}} Category:Group Theory fghfsm379raiteyy6tp8lhpqp8i5qbl
Proof by induction: For all $n \in \N$, let $P \left({n}\right)$ be the proposition $\forall g \in G, h \in H: \left({g, h}\right)^n = \left({g^n, h^n}\right)$. === Basis for the Induction === $P(0)$ is true, as this says: : $\left({g, h}\right)^0 = \left({e_G, e_H}\right)$ $P(1)$ is true, as this says: : $\left({g, h}\right) = \left({g, h}\right)$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\left({g, h}\right)^k = \left({g^k, h^k}\right)$ Then we need to show: :$\left({g, h}\right)^{k+1} = \left({g^{k+1}, h^{k+1}}\right)$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l=\left({g, h}\right)^{k+1}       | r=\left({g, h}\right)^k \circ \left({g, h}\right)       | c= }} {{eqn | r=\left({g^k, h^k}\right) \circ \left({g, h}\right)       | c=Induction Hypothesis }} {{eqn | r=\left({g^k \circ_1 g, h^k \circ_2 h}\right)       | c=by definition of external or internal group product }} {{eqn | r=\left({g^{k+1}, h^{k+1} }\right)       | c= }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: \forall g \in G, h \in H: \left({g, h}\right)^n = \left({g^n, h^n}\right)$ So we have shown the result holds true for all $n \ge 0$. The result for $n < 0$ follows directly from Powers of Group Elements for negative indices. {{qed}} Category:Group Theory Category:Direct Products hsrdqb9vr937yff7dzku2el3m4ojkc6
$\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$ can be proved by induction. For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$ $P(1)$ is true, as this just says: :$\map \phi {s_1} = \map \phi {s_1}$. === Basis for the Induction === $P(2)$ is the case: :$\map \phi {s_1 \circ s_2} = \map \phi {s_1} * \map \phi {s_2}$ This follows from the fact that $\phi$ is a homomorphism. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_k} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_k}$ Then we need to show: :$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_k \circ s_{k + 1} } = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_k} * \map \phi {s_{k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \map \phi {s_1 \circ s_2 \circ \cdots \circ s_k \circ s_{k + 1} }       | r = \map \phi {s_1 \circ s_2 \circ \cdots \circ s_k} * \map \phi {s_{k + 1} }       | c = Basis for the Induction }} {{eqn | r = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_k} * \map \phi {s_{k + 1} }       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$ The result for $n \in \N_{>0}$ follows directly from the above, by replacing each occurrence of $s_k$ with $s$. {{Qed}}
The result for $n \in \N_{>0}$ follows directly from General Morphism Property for Semigroups. For $n = 0$, we use Homomorphism with Cancellable Codomain Preserves Identity. For $n < 0$, we use Homomorphism with Identity Preserves Inverses, along with Index Laws for Monoids: Negative Index. {{qed}}
=== Existence === First, we prove that such a subsemigroup exists. Let $\mathbb S$ be the set of all subsemigroups of $S$ which contain $X$. $\mathbb S \ne \O$ because $S$ is itself a subsemigroup of $S$ (trivial), and thus $S \in \mathbb S$. Let $T$ be the intersection of all the elements of $\mathbb S$. By Intersection of Subsemigroups, $T$ is the largest element of $\mathbb S$ contained in each element of $\mathbb S$. Thus $T$ is a subsemigroup of $S$. Since $\forall x \in \mathbb S: X \subseteq x$, we see that $X \subseteq T$, so $T \in \mathbb S$. === Smallest === Now to show that $T$ is the smallest such subsemigroup. If any $K \le S: X \subseteq K$, then $K \in \mathbb S$ and therefore $T \subseteq K$. So $T$ is the smallest subsemigroup of $S$ containing $X$. === Uniqueness === Now we show that $T$ is unique. Suppose $\exists T_1, T_2 \in \mathbb S$ such that $T_1$ and $T_2$ were two such smallest subsemigroups containing $X$. Then, by the definition of '''smallest''', each would be equal in size. {{WLOG}}, suppose that $T_1$ is not a subset of $T_2$. Then their intersection (by definition containing $X$) would be a subsemigroup smaller than $T_1$. Hence $T_1$ would not be the smallest. Therefore each must be the subset of each other. By definition of set equality, they must be the same set. So the smallest subsemigroup, whose existence we have proved above, is unique. {{Qed}}
Let $H$ be a finite subset of $G$ such that $a, b \in H \implies a \circ b \in H$. From the Two-Step Subgroup Test, it follows that we only need to show that $a \in H \implies a^{-1} \in H$. So, let $a \in H$. First it is straightforward to show by induction that $\set {x \in G: x = a^n: n \in \N} \subseteq H$. That is, $a \in H \implies \forall n \in \N: a^n \in H$. Now, since $H$ is finite, we have that the order of $a$ is finite. Let the order of $a$ be $m$. From Inverse Element is Power of Order Less 1 we have that $a^{m-1} = a^{-1}$. As $a^{m-1} \in H$ (from above) the result follows. {{qed}}
=== Sufficient Condition === Let $H$ be a subgroup of $G$. Then: :$\forall a, b \in H: a \circ b \in H$ by definition of subgroup. {{qed|lemma}} === Necessary Condition === Let $H$ be a non-empty finite subset of $G$ such that: :$\forall a, b \in H: a \circ b \in H$ Let $x \in H$. We have by hypothesis that $H$ is closed under $\circ$. Thus all elements of $\set {x, x^2, x^3, \ldots}$ are in $H$. But $H$ is finite. Therefore it must be the case that: :$\exists r, s \in \N: x^r = x^s$ for $r < s$. So we can write: {{begin-eqn}} {{eqn | l = x^r       | r = x^s       | c =  }} {{eqn | ll= \leadsto       | l = x^r \circ e       | r = x^r \circ x^{s - r}       | c = {{Defof|Identity Element}}, Powers of Group Elements: Sum of Indices }} {{eqn | n = 1       | ll= \leadsto       | l = e       | r = x^{s - r}       | c = Cancellation Laws }} {{eqn | ll= \leadsto       | l = e       | o = \in       | r = H       | c = as $H$ is closed under $\circ$ }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = e       | r = x^{s - r}       | c = which is $(1)$ }} {{eqn | ll= \leadsto       | l = e       | r = x \circ x^{s - r - 1}       | c = as $H$ is closed under $\circ$ }} {{eqn | ll= \leadsto       | l = x^{-1} \circ e       | r = x^{-1} \circ x \circ x^{s - r - 1}       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = x^{-1}       | r = x^{s - r - 1}       | c = {{Defof|Inverse Element}}, {{Defof|Identity Element}} }} {{end-eqn}} But we have that: {{begin-eqn}} {{eqn | l = r       | o = <       | r = s       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = s - r       | o = <       | r = 0 }} {{eqn | ll= \leadsto       | l = s - r - 1       | o = \le       | r = 0 }} {{eqn | ll= \leadsto       | l = x^{s - r - 1}       | o = \in       | r = \set {e, x, x^2, x^3, \ldots}       | c =  }} {{eqn | ll= \leadsto       | l = x^{s - r - 1}       | o = \in       | r = H       | c = as all elements of $\set {e, x, x^2, x^3, \ldots}$ are in $H$ }} {{end-eqn}} So from $(2)$: :$x^{-1} = x^{s - r - 1}$ it follows that: : $x^{-1} \in H$ and from the Two-Step Subgroup Test it follows that $H$ is a subgroup of $G$. {{qed}}
Clearly $a \in H$, so $H \ne \O$. Let $x, y \in H$. {{begin-eqn}} {{eqn | o =        | r = x, y \in H       | c =  }} {{eqn | o = \leadsto       | r = \exists m, n \in \Z: x = a^m, y = a^n       | c =  }} {{eqn | o = \leadsto       | r = x^{-1} y = \paren {a^m}^{-1} a^n       | c =  }} {{eqn | o = \leadsto       | r = x^{-1} y = a^{-m} a^n = a^{n-m}       | c =  }} {{eqn | o = \leadsto       | r = x^{-1} y \in H       | c =  }} {{end-eqn}} Thus by the One-Step Subgroup Test: : $H \le G$ {{qed}}
=== Existence === First, we prove that such a subgroup exists. Let $\mathbb S$ be the set of all subgroups of $G$ which contain $S$. $\mathbb S \ne \O$ because $G$ is itself a subgroup of $G$, and thus $G \in \mathbb S$. Let $H$ be the intersection of all the elements of $\mathbb S$. By Intersection of Subgroups is Subgroup, $H$ is the largest element of $\mathbb S$ contained in each element of $\mathbb S$. Thus $H$ is a subgroup of $G$. Since $\forall x \in \mathbb S: S \subseteq x$, we see that $S \subseteq H$, so $H \in \mathbb S$. === Smallest === Now to show that $H$ is the smallest such subgroup. If any $K \le G: S \subseteq K$, then $K \in \mathbb S$ and therefore $H \subseteq K$. So $H$ is the smallest subgroup of $G$ containing $S$. === Uniqueness === Now we show that $H$ is unique. Suppose $\exists H_1, H_2 \in \mathbb S$ such that $H_1$ and $H_2$ were two such smallest subgroups containing $S$. Then, by the definition of "smallest", each would be equal in size. If one is not a subset of the other, then their intersection (by definition containing $S$) would be a smaller subgroup and hence neither $H_1$ nor $H_2$ would be the smallest. Hence one must be a subset of the other. By definition of set equality, that means they must be the same set. So the smallest subgroup, whose existence we have proved above, is unique. {{Qed}}
Let $H = \left\{{x \in G: \phi \left({x}\right) = \psi \left({x}\right)}\right\}$. From Elements of Group with Equal Images under Homomorphisms form Subgroup, $H$ is a subgroup of $G$. But from the definition of the group generated by $S$, the smallest subgroup that contains $S$ is $G$ itself. Thus $G = \left\{{x \in G: \phi \left({x}\right) = \psi \left({x}\right)}\right\}$ and so $\phi = \psi$. {{Qed}}
Let $H = \gen S$ where $S \subseteq G$. $H$ must certainly include $\hat S$, because any group containing $s \in S$ must also contain $s^{-1}$. Thus $\hat S \subseteq H$. By the closure axiom, $H$ must also contain all products of a finite number of elements of $\hat S$. Thus $\map W {\hat S} \subseteq H$. Now we prove that $\map W {\hat S} \le G$. By the Two-Step Subgroup Test: Let $x, y \in \map W {\hat S}$. 	 As $x$ and $y$ are both products of a finite number of elements of $\hat S$, it follows that so is their product $x y$. Thus $x y \in \map W {\hat S}$ and {{GroupAxiom|0}} is satisfied. Let $x = s_1 s_2 \ldots s_n \in \map W {\hat S}$. Then $x^{-1} = s_n^{-1} \ldots s_2^{-1} s_1^{-1} \in \map W {\hat S}$. Thus the conditions of the Two-Step Subgroup Test are fulfilled, and $\map W {\hat S} \le G$. Thus $\map W {\hat S}$ is the subgroup of $G$ generated by $S$. {{qed}}
It is clear from Set of Words Generates Group that $\map W {\hat X \cup \hat Y} = \gen {X, Y}$. It is equally clear that $X \circ Y \subseteq \map W {\hat X \cup \hat Y}$. {{qed}} Category:Group Theory Category:Subset Products 3cekksmrfte9xtmhqnc5h4td9k4wcc6
From Regular Representations of Subset Product, we have that the left regular representation of $\struct {S, \circ}$ with respect to $a$ is: :$\lambda_x \sqbrk S = \set x \circ S = x \circ S$ The result then follows directly from Regular Representation of Invertible Element is Permutation. {{qed}}
Consider the relation $\RR \subseteq G \times G$ defined as: :$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists g \in X$ Then: :$\forall S \subseteq G: X \circ S = \RR \sqbrk S$ Then: {{begin-eqn}} {{eqn | l = X \circ \paren {Y \cap Z}       | r = \RR \sqbrk {Y \cap Z}       | c =  }} {{eqn | o = \subseteq       | r = \RR \sqbrk Y \cap \RR \sqbrk Z       | c = Image of Intersection under Relation }} {{eqn | r = \paren {X \circ Y} \cap \paren {X \circ Z}       | c =  }} {{end-eqn}} Next, consider the relation $\RR \subseteq G \times G$ defined as: :$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists h \in X$ Then: :$\forall S \subseteq G: S \circ X = \RR \sqbrk S$ Then: {{begin-eqn}} {{eqn | l = \paren {Y \cap Z} \circ X       | r = \RR \sqbrk {Y \cap Z}       | c =  }} {{eqn | o = \subseteq       | r = \RR \sqbrk Y \cap \RR \sqbrk Z       | c = Image of Intersection under Relation }} {{eqn | r = \paren {Y \circ X} \cap \paren {Z \circ X}       | c =  }} {{end-eqn}} {{qed}}
Let $x \in X, t \in Y \cap Z$. By the definition of intersection, $t \in Y$ and $t \in Z$. Consider $X \circ \paren {Y \cap Z}$. We have $x \circ t \in X \circ \paren {Y \cap Z}$ by definition of subset product. As $t \in Y$ and $t \in Z$, we also have $x \circ t \in X \circ Y$ and $x \circ t \in X \circ Z$. The result follows. Similarly, consider $\paren {Y \cap Z} \circ X$. Then we have $t \circ x \in \paren {Y \cap Z} \circ X$ by definition of subset product. As $t \in Y$ and $t \in Z$, we also have $t \circ x \in Y \circ X$ and $t \circ x \in Z \circ X$. Again, the result follows. {{Qed}}
From Intersection of Subgroups is Subgroup, we have that $H \cap K \le H$. Let the number of left cosets of $H \cap K$ in $H$ be $r$. Then the left coset space of $H \cap K$ in $H$ is: : $\set {x_1 \paren {H \cap K}, x_2 \paren {H \cap K}, \ldots, x_r \paren {H \cap K} }$ So each element of $H$ is in $x_i \paren {H \cap K}$ for some $1 \le i \le r$. Also, if $i \ne j$, we have: :$x_j^{-1} x_i \notin H \cap K$ Let $h k \in H K$. We can write $h = x_i g$ for some $1 \le i \le r$ and some $g \in K$. Thus: :$h k = x_i \paren {g k}$ Since $g, k \in K$, this shows $h k \in x_i K$. {{AimForCont}} the left cosets $x_i K$ are not all disjoint. Then by Left Coset Space forms Partition: :$x_i K = x_j K$ for some $i, j$. So by Left Congruence Class Modulo Subgroup is Left Coset: :$x_j^{-1} x_i \in K$ Since $x_i, x_j \in H$, we have: :$x_j^{-1} x_i \in H \cap K$ which contradicts the definition. Therefore the left cosets $x_i K$ are disjoint for $1 \le i \le r$. This leads us to: :$\dfrac {\order H} {\order {H \cap K} } = \dfrac {\order {H K} } {\order K} = r$ whence the result. {{qed}}