## The set of all endomorphisms of an abelian group is a ring under pointwise addition and composition

Let be any abelian group. Let be the set of all group homomorphisms of the additive group to itself, with addition defined pointwise and with multiplication defined as function composition. Prove that these operations make into a ring with identity. Prove that the units of are precisely the group automorphisms of .

Let , and let .

- Note that . Thus , so that addition is associative.
- Consider the trivial homomorphism . If is a homomorphism , then . Thus . Similarly, , so that this set has an additive identity.
- Define the mapping . Since , is a group homomorphism. Moreover, , so that . Similarly, . Thus every element has an additive inverse.
- Note that , so that , and is an abelian group under addition.
- Recall that function composition is always associative, so that multiplication is associative.
- Note that . Thus , and multiplication distributes over addition on the left.
- Similarly, , and we have . Thus multiplication distributes over addition on the right.
- Finally, let denote the identity homomorphism. Clearly , so that 1 is a multiplicative identity element.
- If , then . If , then clearly . That is, if and only if is nontrivial.

Thus is a ring with 1, and if and only if is nontrivial.

Now we consider the units in .

If is a unit, then there exists a homomorphism such that . Thus is injective and surjective, and in fact is an automorphism of . Conversely, if is an automorphism of , then is an automorphism, and we have . Thus is a unit. That is, the units of are precisely the group automorphisms of .

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