Endomorphism Ring of Abelian Group is Ring with Unity

== Theorem ==
Let $\struct {G, +}$ be an abelian group.
Let $\struct {\map {\mathrm {End} } G, \oplus, \circ}$ be its endomorphism ring.
Then $\struct {\map {\mathrm {End} } G, \oplus, \circ}$ is a ring with unity $I_G$, where:
:$I_G$ is the identity mapping on $G$
:$\oplus$ is the pointwise operation on $\map {\mathrm {End} } G$ induced by $+$.
== Proof ==
In the below, $f, g , h \in \map {\mathrm {End} } G$.
As $f, g, h$ are (group) endomorphisms, it follows that they are also (group) homomorphisms.
Recall that $f \oplus g$ is the pointwise operation on $f$ and $g$ induced by $+$.
We are given that $\struct {G, +}$ is an abelian group.
Hence $\struct {G, +}$ is a commutative semigroup.
By Homomorphism on Induced Structure to Commutative Semigroup, $f \oplus g$ is a homomorphism from $\struct {G, +}$ into $\struct {G, +}$.
So, by definition, $f \oplus g$ is an endomorphism on $\struct {G, +}$.
Now, taking the ring axioms in turn:
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We have that $f \oplus g$ is an endomorphism on $\struct {G, +}$.
Hence:
:$\forall f, g \in \map {\mathrm {End} } G: f \oplus g \in \map {\mathrm {End} } G$
Hence, by definition, $\struct {\map {\mathrm {End} } G, \oplus}$ is closed.
=== ===
By Homomorphism on Induced Structure to Commutative Semigroup again, we have:
:$\forall f, g, h \in \map {\mathrm {End} } G: \paren {f \oplus g} \oplus h = f \oplus \paren {g \oplus h}$
Hence $\oplus$ is associative on $\map {\mathrm {End} } G$.
=== ===
By Homomorphism on Induced Structure to Commutative Semigroup again, we have:
:$\forall f, g \in \map {\mathrm {End} } G:\paren{f \oplus g} = \paren{g \oplus f}$
Hence $\oplus$ is commutative on $\map {\mathrm {End} } G$.
=== ===
We are given that $I_G$ is the identity mapping on $G$.
By Identity Mapping is Ring Automorphism, $I_G$ is an automorphism on $G$.
Hence $I_G$ is an endomorphism on $G$:
:$I_G \in \map {\mathrm {End} } G$
=== ===
Follows from Pointwise Inverse in Induced Structure.
=== ===
By Composite of Homomorphisms is Homomorphism, for any homomorphisms $f$ and $g$, $f \circ g$ is also a homomorphism.
By Composite of Endomorphisms is Endomorphism, if $f$ and $g$ are endomorphisms, then the composite of $f$ and $g$ is also an endomorphism
Therefore:
:$\forall f, g \in \map {\mathrm {End} } G: f \circ g \in \map {\mathrm {End} } G$
Hence $\map {\mathrm {End} } G$ is closed under $\circ$.
=== ===
From Structure Induced by Associative Operation is Associative we have that:
:$\paren {f \circ g} \circ h = f \circ \paren {g \circ h}$
So we have:
:$\forall f, g, h \in \map {\mathrm {End} } G: \paren{f \circ g} \circ h = f \circ \paren{g \circ h}$
So $\circ$ is associative on $\map {\mathrm {End} } G$:
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By Composition of Mappings is Left Distributive over Homomorphism of Pointwise Operation:
:$f \circ \paren {g \oplus h} = \paren {f \oplus g} \circ \paren {f \oplus h}$
By Composition of Mappings is Right Distributive over Pointwise Operation:
:$\paren {g \oplus h} \circ f= \paren {g \oplus f} \circ \paren {h \oplus f}$
So it follows from definition of distributive that composition of mappings is distributive over homomorphism of the pointwise operation $\oplus$.
As $f, g, h$ are endomorphisms, it follows that composition of mappings is distributive over endomorphism of $\oplus$.
Hence $\struct {\map {\mathrm {End} } G, \oplus, \circ}$ is a ring with unity $I_G$.
== Sources ==
* : Chapter $0$: Algebraic Concepts
Category:Abstract Algebra
Category:Rings of Endomorphisms
Category:Endomorphism Rings of Abelian Groups