hcju/proofwikiqa huggingface dataset:
Let $x \circ' y^{-1}$ be cancellable for $\circ'$, where $x \in S$ and $y \in C$. We have that $y$ is invertible for $\circ'$. So by Invertible Element of Associative Structure is Cancellable: :$y$ is cancellable for $\circ'$. Now by definition of inverse element: :$x = \left({x \circ' y^{-1}}\right) \circ' y$ Thus $x$ is also cancellable for $\circ'$. By Cancellable Elements of Semigroup form Subsemigroup, $x$ is cancellable for $\circ$. So: : $x \in S \implies x \in C$ Thus $x$ is invertible for $\circ'$. Hence by Inverse of Product in Associative Structure: : $x \circ' y$ is invertible for $\circ'$. So every cancellable element of $\left({T, \circ'}\right)$ is invertible. So, by definition, $\left({T, \circ'}\right)$ is its own inverse completion. {{qed}}
Let $e$ be the identity for $\circ$. Let $e = x \circ' y^{-1}$, where $x \in S, y \in C$. Then: {{begin-eqn}} {{eqn | l = y       | r = e \circ y       | c = Definition of Identity }} {{eqn | r = \left({x \circ' y^{-1} }\right) \circ' y       | c = Definition of Inverse }} {{eqn | r = x \circ' \left({y^{-1} \circ' y}\right)       | c = $\circ'$ is associative }} {{eqn | r = x \circ' e       | c = Definition of Inverse }} {{end-eqn}} Thus $e = y^{-1} \circ' y$, and $y^{-1} \circ' y$ is the identity for $\circ'$. {{qed}}
Let $\struct {S, \circ}$ be a commutative semigroup, all of whose elements are cancellable. Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$. From Inverse Completion is Commutative Monoid: :$\struct {T, \circ'} = \struct {S \circ' S^{-1}, \circ'}$ has been shown to be a commutative monoid. Taking the group axioms in turn: === {{GroupAxiom|0|nolink}} === As $\struct {T, \circ'}$ is a commutative semigroup, it is by definition closed. {{qed|lemma}} === {{GroupAxiom|1|nolink}} === As $\struct {T, \circ'}$ is a commutative semigroup, it is by definition associative. {{qed|lemma}} === {{GroupAxiom|2|nolink}} === Let $x \in S$. Then $x^{-1} \in S^{-1}$ by definition. As $\struct {T, \circ'} = \struct {S \circ' S^{-1}, \circ'}$ is closed: :$x \circ' x^{-1} \in T$ This holds for all $x \in S$. As $\struct {T, \circ'}$ is a commutative semigroup: :$\exists e \in T: \forall x \in S: x \circ' x^{-1} = e^{-1} = x \circ' x$ Thus $\struct {T, \circ'}$ has an identity element. {{qed|lemma}} === {{GroupAxiom|3|nolink}} === Every element of $S$ has an inverse in $S^{-1}$. Therefore every element of $S$ and $S^{-1}$ is invertible. From Inverse of Product in Associative Structure, every element of $S \circ' S^{-1}$ is therefore also invertible. Thus every element of $T$ has an inverse. {{qed|lemma}} All the group axioms are thus seen to be fulfilled, and so $\struct {T, \circ'}$ is a group. Commutativity of $\circ$ has been demonstrated. Hence the result, by definition of abelian group. {{qed}}
Let $\left({S, \circ}\right)$ be a commutative semigroup which has cancellable elements. From Construction of Inverse Completion, we can construct an inverse completion $\left({T', \oplus'}\right)$ of $\left({S', \oplus'}\right)$, which is an isomorphic copy of $S$ under the mapping $\psi: S \to S'$. By the Embedding Theorem, there exists a semigroup $\left({T, \circ}\right)$  which contains $\left({S, \circ}\right)$, and an isomorphism $\Psi$ from $\left({T, \circ}\right)$ to $\left({T', \oplus'}\right)$ which extends $\psi$. Thus $T$ is an inverse completion of $S$. {{Qed}}
Let $C$ be the subsemigroup of cancellable elements of $S$. It is proved that this is a semigroup by Cancellable Elements of Semigroup form Subsemigroup. The set of cancellable elements of $T$ is $\phi \left({C}\right)$. By the Extension Theorem for Homomorphisms, there is: : A unique homomorphism $\phi'$ from $S'$ into $T'$ extending $\phi$ and: : A unique homomorphism $\psi$ from $T'$ into $S'$ extending $\phi^{-1}$. Therefore, $\psi \circ \phi'$ is an endomorphism of $S'$ whose restriction to $S$ is the inclusion mapping $\iota_S: S \to S'$: :$\forall x \in S: \iota \left({x}\right) = x$ From Inclusion Mapping is Monomorphism, $\iota_S$ is a monomorphism from $S$ into $S'$. But by the Extension Theorem for Homomorphisms, the identity automorphism of $S'$ is the only endomorphism of $S'$ extending $\iota_S$ from $S$ into $S'$, so: :$\psi \circ \phi' = I_{S'}$ Similarly: :$\phi' \circ \psi = I_{T'}$ So by Bijection iff Left and Right Inverse, $\phi'$ is a bijection and therefore an isomorphism. {{Qed}}
Let $T$ and $T'$ both be inverse completions of a commutative semigroup $S$ having cancellable elements. Then from the Extension Theorem for Isomorphisms, there is a unique isomorphism $\phi: T \to T'$ satisfying $\forall x \in S: \phi \left({x}\right) = x$. Hence the result. {{Qed}}
Let $\left({S, \circ}\right)$ be a the monoid. By definition: :Identity: $\exists e_S \in S: \forall a \in S: a \circ e_S = a = e_S \circ a$ So a monoid must ''at least'' have an identity. Therefore $e_S \in S$ and so $S$ is not the empty set. {{qed}}
A group is defined as a monoid for which every element has an inverse. Thus, as a group is already a monoid, it must ''at least'' have an identity, therefore can not be empty. {{qed}}
The Identity Element is Idempotent. From the Cancellation Laws, all group elements are cancellable. The result follows from Identity is only Idempotent Cancellable Element. {{qed}}
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $x \in G$ such that $x \circ x = x$. {{begin-eqn}} {{eqn | l = e       | r = x \circ x^{-1}       | c = {{GroupAxiom|3}} }} {{eqn | r = \paren {x \circ x} \circ x^{-1}       | c = by hypothesis: $x \circ x = x$ }} {{eqn | r = x \circ \paren {x \circ x^{-1} }       | c = {{GroupAxiom|1}} }} {{eqn | r = x \circ e       | c = {{GroupAxiom|3}} }} {{eqn | r = x       | c = {{GroupAxiom|2}} }} {{end-eqn}} {{qed}}
Let the identity element of $\struct {G, \circ}$ be $e_G$. === Necessary Condition === Let $x$ and $y$ commute. Then: {{begin-eqn}} {{eqn | l = \paren {x \circ y} \circ \paren {x \circ y}       | r = x \circ \paren {y \circ x} \circ y       | c = $\circ$ is associative }} {{eqn | r = x \circ \paren {x \circ y} \circ y       | c = $x$ and $y$ commute }} {{eqn | r = \paren {x \circ x} \circ \paren  {y \circ y}       | c = $\circ$ is associative }} {{eqn | r = e_G \circ e_G       | c = $x$ and $y$ are self-inverse }} {{eqn | r = e_G       | c = {{Defof|Identity Element}} }} {{end-eqn}} Thus $\paren {x \circ y} \circ \paren {x \circ y} = e_G$, proving that $x \circ y$ is self-inverse. {{qed|lemma}} === Sufficient Condition === Now, suppose that $x \circ y$ is self-inverse. We already have that $x$ and $y$ are self-inverse. Thus: :$\paren {x \circ x} \circ \paren {y \circ y} = e_G \circ e_G = e_G$ Because $x \circ y$ is self-inverse, we have: :$\paren {x \circ y} \circ \paren {x \circ y} = e_G$ Thus: {{begin-eqn}} {{eqn | l = \paren {x \circ y} \circ \paren {x \circ y}       | r = \paren {x \circ x} \circ \paren {y \circ y} = e_G       | c =  }} {{eqn | ll= \leadsto       | l = x \circ \paren {y \circ x} \circ y       | r = x \circ \paren {x \circ y} \circ y       | c = $\circ$ is associative }} {{eqn | ll= \leadsto       | l = y \circ x       | r = x \circ y       | c = Cancellation Laws }} {{end-eqn}} So $x$ and $y$ commute. {{qed}}
We need to prove closure and associativity. === Closure === Let $\struct {G, \circ}$ be a group, and let $A, B \subseteq G$. {{begin-eqn}} {{eqn | o =        | r = \forall a \in A, b \in B: a \circ b \in G       | c =  }} {{eqn | o = \leadsto       | r = A \circ B \subseteq G       | c = {{Defof|Subset Product}} }} {{eqn | o = \leadsto       | r = A \circ B \in \powerset G       | c = {{Defof|Power Set}} }} {{end-eqn}} Thus $\struct {\powerset G, \circ_\PP}$ is closed. {{qed|lemma}} === Associativity === It follows from Subset Product within Semigroup is Associative that $\circ_\PP$ is associative in $\struct {\powerset G, \circ_\PP}$. {{qed|lemma}} Thus $\struct {\powerset G, \circ_\PP}$ is a semigroup. {{qed}}
By definition a group is also a semigroup. The result then follows from Power Set of Semigroup under Induced Operation is Semigroup. {{qed}}
This follows directly from the fact that all elements of a group are by definition invertible. Therefore the result Regular Representation of Invertible Element is Permutation applies. {{Qed}}
Follows immediately from the fact that both the left and right regular representation are permutations, and therefore bijections. {{qed}}
Every element of $G$ is self-inverse, that is: :$\forall x \in G: x \circ x = e$ In particular, for all $x, y \in G$: :$\paren {x \circ y} \circ \paren {x \circ y} = e$ that is, $x \circ y$ is also self-inverse. From Self-Inverse Elements Commute iff Product is Self-Inverse, it follows that: :$\forall x, y \in G: y \circ x = x \circ y$ that is: $x$ and $y$ commute for all $x, y \in G$. Hence $G$ is an abelian group. {{qed}}
{{begin-eqn}} {{eqn | l = x \circ y       | r = y \circ x }} {{eqn | ll= \iff       | l = \paren {x \circ y} \circ x^{-1}       | r = \paren {y \circ x} \circ x^{-1}       | c = Cancellation Laws }} {{eqn | ll= \iff       | l = x \circ y \circ x^{-1}       | r = y \circ \paren {x \circ x^{-1} }       | c = {{Defof|Associative Operation}} }} {{eqn | ll= \iff       | l = x \circ y \circ x^{-1}       | r = y \circ e       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \iff       | l = x \circ y \circ x^{-1}       | r = y       | c = {{Defof|Identity Element}} }} {{end-eqn}} {{qed}}
By definition, an automorphism is an isomorphism from an algebraic structure onto itself. An isomorphism, in turn, is a bijective homomorphism. From Identity Mapping is Bijection, the identity mapping $I_S: S \to S$ on the set $S$ is a bijection from $S$ onto itself. Now we need to show it is a homomorphism. Let $x, y \in S$. Then: {{begin-eqn}} {{eqn | l=I_S \left({x \circ y}\right)       | r=x \circ y       | c=Definition of Identity Mapping }} {{eqn | r=I_S \left({x}\right) \circ I_S \left({y}\right)       | c=Definition of Identity Mapping }} {{end-eqn}} Thus $I_S \left({x \circ y}\right) = I_S \left({x}\right) \circ I_S \left({y}\right)$ and the morphism property holds, proving that $I_S: S \to S$ is a homomorphism. As $I_S$ is a bijection, its image is $S$. {{Qed}}
=== Necessary Condition === Let $\struct {G, \circ}$ be an abelian group. Let $a, b \in G$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = \map \phi {a \circ b}       | r = \paren {a \circ b} \circ \paren {a \circ b}       | c = Definition of $\phi$ }} {{eqn | r = a \circ \paren {b \circ a} \circ b       | c = {{GroupAxiom|1}} }} {{eqn | r = a \circ \paren {a \circ b} \circ b       | c = {{Defof|Abelian Group}}: Commutativity }} {{eqn | r = \paren {a \circ a} \circ \paren {b \circ b}       | c = {{GroupAxiom|1}} }} {{eqn | r = \map \phi a \circ \map \phi b       | c = Definition of $\phi$ }} {{end-eqn}} As $a$ and $b$ are arbitrary, the above holds for all $a, b \in G$. Thus $\phi$ is a group homomorphism from $G$ to $G$. So by definition, $\phi$ is a group endomorphism. {{qed|lemma}} === Sufficient Condition === Let $\phi: G \to G$ as defined above be a group endomorphism. Then: {{begin-eqn}} {{eqn | lo= \forall a, b \in G:       | l = \map \phi {a \circ b}       | r = \map \phi a \circ \map \phi b       | c = {{Defof|Group Endomorphism}} }} {{eqn | ll= \leadsto       | lo= \forall a, b \in G:       | l = \paren {a \circ b} \circ \paren {a \circ b}       | r = \paren {a \circ a} \circ \paren {b \circ b}       | c = Definition of $\phi$ }} {{eqn | ll= \leadsto       | lo= \forall a, b \in G:       | l = a \circ \paren {b \circ a} \circ b       | r = a \circ \paren {a \circ b} \circ b       | c = {{GroupAxiom|1}} }} {{eqn | ll= \leadsto       | lo= \forall a, b \in G:       | l = b \circ a       | r = a \circ b       | c = Cancellation Laws }} {{end-eqn}} Thus, by definition, $G$ is an abelian group. {{qed}}
We have $\tuple {h_1, h_2} \circ \tuple {k_1, k_2} = \tuple {h_1 \circ k_1, h_2 \circ k_2}$ by definition of group direct product. === Necessary Condition === Let $\phi$ be a homomorphism. Then: {{begin-eqn}} {{eqn | l = \map \phi {\tuple {h_1, h_2} \circ \tuple {k_1, k_2} }       | r = \map \phi {h_1, h_2} \circ \map \phi {k_1, k_2}       | c =  }} {{eqn | ll= \leadsto       | l = \map \phi {h_1 \circ k_1, h_2 \circ k_2}       | r = \tuple {h_1 \circ h_2} \circ \tuple {k_1 \circ k_2}       | c =  }} {{eqn | ll= \leadsto       | l = h_1 \circ k_1 \circ h_2 \circ k_2       | r = h_1 \circ h_2 \circ k_1 \circ k_2       | c =  }} {{eqn | ll= \leadsto       | l = k_1 \circ h_2       | r = h_2 \circ k_1       | c =  }} {{end-eqn}} This follows whatever $k_1$ and $h_2$ are. So in order for $\phi$ to be a homomorphism, every element of $H_1$ must commute with every element of $H_2$. {{qed|lemma}} === Sufficient Condition === Let every element of $H_1$ commute with every element of $H_2$. Let $\tuple {h_1, h_2}, \tuple {k_1, k_2} \in H_1 \times H_2$. Then: {{begin-eqn}} {{eqn | l = \map \phi {\tuple {h_1, h_2} \circ \tuple {k_1, k_2} }       | r = \map \phi {h_1 \circ k_1, h_2 \circ k_2}       | c =  }} {{eqn | r = h_1 \circ k_1 \circ h_2 \circ k_2       | c =  }} {{eqn | r = h_1 \circ h_2 \circ k_1 \circ k_2       | c = $k_1$ commutes with $h_2$ }} {{eqn | r = \map \phi {h_1, h_2} \circ \map \phi {k_1, k_2}       | c =  }} {{end-eqn}} Thus $\phi$ is shown to be a homomorphism. {{Qed}} Category:Group Homomorphisms ahllokut0j0saikd89t07n91h4lhxgg
We have that Isomorphism Preserves Commutativity. Thus: :$\forall x, y \in G: x \circ y = y \circ x \implies \map \phi x * \map \phi y = \map \phi y * \map \phi x$ Thus if $G$ is abelian, so is $H$. As $\phi^{-1}: H \to G$ is also an isomorphism, it is clear that if $H$ is abelian, then so is $G$. {{qed}}
=== $\text G 0$: Closure === $\struct {G, *}$ is closed: :$b \circ a \in G \implies a * b \in G$ {{qed|lemma}} === $\text G 1$: Associativity === $*$ is associative on $G$: {{begin-eqn}} {{eqn | l = a * \paren {b * c}       | r = \paren {c \circ b} \circ a       | c = Definition of $*$ }} {{eqn | r = c \circ \paren {b \circ a}       | c = Associativity of $\circ$ }} {{eqn | r = \paren {a * b} * c       | c = Definition of $*$ }} {{end-eqn}} {{qed|lemma}} === $\text G 2$: Identity === Let $e$ be the identity of $\struct {G, \circ}$: {{begin-eqn}} {{eqn | l = a * e       | r = e \circ a = a }} {{eqn | l = e * a       | r = a \circ e = a }} {{end-eqn}} Thus $e$ is the identity of $\struct {G, *}$. {{qed|lemma}} === $\text G 3$: Inverses === Let the inverse of $a \in \struct {G, \circ}$ be $a^{-1}$: {{begin-eqn}} {{eqn | l = a * a^{-1}       | r = a^{-1} \circ a = e }} {{eqn | l = a^{-1} * a       | r = a \circ a^{-1} = e }} {{end-eqn}} Thus $a^{-1}$ is the inverse of $a \in \struct {G, *}$ {{qed|lemma}} So all the group axioms are satisfied, and $\struct {G, *}$ is a group. {{qed}}
=== $G \, 0$: Closure === Let $x, y \in G$. Then: :$\forall x * y = x \circ c^{-1} \circ y \in G$ as $c^{-1} \in G$ thus demonstrating that $\struct {G, *}$ is closed. {{qed|lemma}} === $G \, 1$: Associativity === Let $x, y, z \in G$. {{begin-eqn}} {{eqn | l = x * \paren {y * z}       | r = x \circ c^{-1} \circ \paren {y \circ c^{-1} \circ z}       | c = Definition of $*$ }} {{eqn | r = \paren {x \circ c^{-1} \circ y} \circ c^{-1} \circ z       | c = Associativity of $\circ$ }} {{eqn | r = \paren {x * y} * z       | c = Definition of $*$ }} {{end-eqn}} thus demonstrating that $\struct {G, *}$ is associative. {{qed|lemma}} === $G \, 2$: Identity === Let $x \in G$. {{begin-eqn}} {{eqn | l = x * c       | r = x \circ c^{-1} \circ c = x }} {{eqn | l = c * x       | r = c \circ c^{-1} \circ x = x }} {{end-eqn}} So $c$ serves as the identity in $\struct {G, *}$. {{qed|lemma}} === $G \, 3$: Inverses === Let $x \in G$. We need to find $y \in G$ such that $x * y = c \implies x \circ c^{-1} \circ y = c$. {{begin-eqn}} {{eqn | l = x * y       | r = c       | c =  }} {{eqn | ll= \leadsto       | l = x \circ c^{-1} \circ y       | r = c       | c =  }} {{eqn | ll= \leadsto       | l = c^{-1} \circ y       | r = x^{-1} \circ c       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = c \circ x^{-1} \circ c       | c =  }} {{end-eqn}} Thus the inverse of $x$ under the operation $*$ is $c \circ x^{-1} \circ c$ where $x^{-1}$ is the inverse of $x$ under $\circ$. {{qed|lemma}} All criteria have been demonstrated to be fulfilled, and so $\struct {G, *}$ is a group. {{qed}}
From Power Set is Closed under Symmetric Difference, we have that $\struct {\powerset S, *}$ is closed. The result follows directly from Set System Closed under Symmetric Difference is Abelian Group. {{Qed}}
By Set is Subset of Itself, we have that: :$G \subseteq G$ Thus $\struct {G, \circ}$ is a group which is a subset of $\struct {G, \circ}$, and therefore a subgroup of $\struct {G, \circ}$. {{Qed}}
From the definition, a group is a monoid. Also, all of its elements are cancellable. The result then follows directly from the result for monoids each of whose elements are cancellable: Identity of Cancellable Monoid is Identity of Submonoid. {{qed}}
Let $h \in H$. Let: : $h'$ be the inverse of $h$ in $H$ : $h^{-1}$ be the inverse of $h$ in $G$. From Identity of Subgroup, $h \circ h' = e$. From Inverse in Group is Unique, it follows that $h' = h^{-1}$. {{qed}}
Follows directly from Restriction of Commutative Operation is Commutative. {{qed}}
Let $H = H_1 \cap H_2$ where $H_1, H_2 \le \struct {G, \circ}$. Then: {{begin-eqn}} {{eqn | o =        | r = a, b \in H       | c =  }} {{eqn | o = \leadsto       | r = a, b \in H_1 \land a, b \in H_2       | c = {{Defof|Set Intersection}} }} {{eqn | o = \leadsto       | r = a \circ b^{-1} \in H_1 \land a \circ b^{-1} \in H_2       | c = Group Properties }} {{eqn | o = \leadsto       | r = a \circ b^{-1} \in H       | c = {{Defof|Set Intersection}} }} {{eqn | o = \leadsto       | r = H \le G       | c = One-Step Subgroup Test }} {{end-eqn}} As $H \subseteq H_1$ and $H \subseteq H_2$, the other results follow directly. {{qed}}
As neither $H \subseteq K$ nor $K \subseteq H$, it follows from Set Difference with Superset is Empty Set that neither $H \setminus K = \O$ nor $K \setminus H = \O$. So, let $h \in H \setminus K, k \in K \setminus H$. Thus, $h \notin K, k \notin H$. If $\struct {H \cup K, \circ}$ is a group, then it must be closed. If $\struct {H \cup K, \circ}$ is closed, then $h \circ k \in H \cup K \implies h \circ k \in H \lor h \circ k \in K$. If $h \circ k \in H$ then $h^{-1} \circ h \circ k \in H \implies k \in H$. If $h \circ k \in K$ then $h \circ k \circ k^{-1} \in K \implies h \in K$. So $h \circ k$ can be in neither $H$ nor $K$. Therefore $\struct {H \cup K, \circ}$ is not closed. Therefore $H \cup K$ is not a subgroup of $G$. {{qed}}
Let the identities of $\struct {G, \circ}$ and $\struct {H, *}$ be $e_G$ and $e_H$ respectively. By Homomorphism to Group Preserves Identity: :$\map f {e_G} = \map g {e_G} = e_H$ Thus $e_G \in S$, and so $S \ne \O$. Similarly, from Homomorphism to Group Preserves Inverses, $x \in S \implies x^{-1} \in S$. Let $x, y \in S$. Then: {{begin-eqn}} {{eqn | l = \map f {x \circ y}       | r = \map f x * \map f y       | c = Morphism Property }} {{eqn | r = \map g x * \map g y       | c = Definition of $S$ }} {{eqn | r = \map g {x \circ y}       | c = Morphism Property }} {{end-eqn}} Thus $x \circ y \in S$. So, by the Two-Step Subgroup Test: :$S \le G$ {{Qed}}
From Magma Subset Product with Self, we have: :$H \circ H \subseteq H$ Let $e$ be the identity of $G$. By Identity of Subgroup, it is also the identity of $H$. So: {{begin-eqn}} {{eqn | l = h       | o = \in       | r = H       | c =  }} {{eqn | ll= \leadsto       | l = e \circ h       | o = \in       | r = H \circ H       | c =  }} {{eqn | ll= \leadsto       | l = h       | o = \in       | r = H \circ H       | c =  }} {{eqn | ll= \leadsto       | l = H       | o = \subseteq       | r = H \circ H       | c =  }} {{end-eqn}} Hence the result from the definition of set equality. {{qed}}
As $H$ is a subgroup of $G$: :$\forall h \in H: h^{-1} \in H$ The result follows.
Let $H \le G_1$. First note that from Null Relation is Mapping iff Domain is Empty Set: :$H \ne \O \implies \phi \sqbrk H \ne \O$ and so $\phi \sqbrk H$ is not empty. Next, let $x, y \in \phi \sqbrk H$. Then: :$\exists h_1, h_2 \in H: x = \map \phi {h_1}, y = \map \phi {h_2}$ From the definition of Group Homomorphism, we have: :$\map \phi {h_1^{-1} \circ h_2} = x^{-1} * y$ Since $H$ is a subgroup: :$h_1^{-1} \circ h_2 \in H$ Hence: :$x^{-1} * y \in \phi \sqbrk H$ Thus from the One-Step Subgroup Test: :$\phi \sqbrk H \le G_2$ {{qed}}
This is a special case of Group Homomorphism Preserves Subgroups, where we set $H = G_1$. {{Qed}}
Checking each of the criteria for an equivalence relation in turn: === Reflexive === :$\forall x \in G: e_G \circ x = x \circ e_G \implies x \sim x$ Thus conjugacy of group elements is reflexive. {{qed|lemma}} === Symmetric === {{begin-eqn}} {{eqn | o =        | r = x \sim y       | c =  }} {{eqn | o = \leadsto       | r = a \circ x = y \circ a       | c = {{Defof|Conjugate of Group Element}} }} {{eqn | o = \leadsto       | r = a \circ x \circ a^{-1} = y       | c = {{Defof|Group}} }} {{eqn | o = \leadsto       | r = a^{-1} \circ y = x \circ a^{-1}       | c = {{Defof|Group}} }} {{eqn | o = \leadsto       | r = y \sim x       | c = {{Defof|Conjugate of Group Element}} }} {{end-eqn}} Thus conjugacy of group elements is symmetric. {{qed|lemma}} === Transitive === {{begin-eqn}} {{eqn | o =        | r = x \sim y, y \sim z       | c =  }} {{eqn | o = \leadsto       | r = a_1 \circ x = y \circ a_1, a_2 \circ y = z \circ a_2       | c = {{Defof|Conjugate of Group Element}} }} {{eqn | o = \leadsto       | r = a_2 \circ a_1 \circ x = a_2 \circ y \circ a_1 = z \circ a_2 \circ a_1       | c = {{Defof|Group}} }} {{eqn | o = \leadsto       | r = x \sim z       | c = {{Defof|Conjugate of Group Element}} }} {{end-eqn}} Thus conjugacy of group elements is transitive. {{qed|lemma}} All criteria are satisfied, and so conjugacy of group elements is shown to be an equivalence relation. {{Qed}}
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism. From Homomorphism to Group Preserves Identity, $\map \phi {e_G} = e_H$, so $e_G \in \map \ker \phi$. Therefore $\map \ker \phi \ne \O$. Let $x, y \in \map \ker \phi$, so that $\map \phi x = \map \phi y = e_H$. Then: {{begin-eqn}} {{eqn | l = \map \phi {x^{-1} \circ y}       | r = \map \phi {x^{-1} } * \map \phi y       | c = {{Defof|Morphism Property}} }} {{eqn | r = \paren {\map \phi x}^{-1} * \map \phi y       | c = Homomorphism with Identity Preserves Inverses }} {{eqn | r = e_T^{-1} * e_T       | c = as $x, y \in \map \ker \phi$ }} {{eqn | r = e_T       | c = {{Defof|Identity Element}} }} {{end-eqn}} So $x^{-1} \circ y \in \map \ker \phi$, and from the One-Step Subgroup Test, $\map \ker \phi \le S$. {{qed}}
Let $\struct {S, \circ}$ be a commutative semigroup. Let $a, b, c, d \in S$. Then: {{begin-eqn}} {{eqn | l = \paren {a \circ b} \circ \paren {c \circ d}       | r = a \circ \paren {b \circ \paren {c \circ d} }       | c = Associativity of $\circ$ }} {{eqn | r = a \circ \paren {\paren {b \circ c} \circ d}       | c = Associativity of $\circ$ }} {{eqn | r = a \circ \paren {\paren {c \circ b} \circ d}       | c = Commutativity of $\circ$ }} {{eqn | r = a \circ \paren {c \circ \paren {b \circ d} }       | c = Associativity of $\circ$ }} {{eqn | r = \paren {a \circ c} \circ \paren {b \circ d}       | c = Associativity of $\circ$ }} {{end-eqn}} {{qed}}