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

Let A be any abelian group. Let R = \mathsf{Hom}(A,A) be the set of all group homomorphisms of the additive group A to itself, with addition defined pointwise and with multiplication defined as function composition. Prove that these operations make R into a ring with identity. Prove that the units of R are precisely the group automorphisms of A.


Let a,b,c \in R, and let x \in A.

  1. Note that (a+(b+c))(x) = a(x) + (b+c)(x) = a(x) + (b(x) + c(x)) = (a(x) + b(x)) + c(x) = (a+b)(x) + c(x) = ((a+b)+c)(x). Thus a+(b+c) = (a+b)+c, so that addition is associative.
  2. Consider the trivial homomorphism 0(x) = 0. If a is a homomorphism A \rightarrow A, then (0 + a)(x) = 0(x) + a(x) = a(x). Thus 0+a = a. Similarly, a + 0 = a, so that this set has an additive identity.
  3. Define the mapping \overline{a}(x) = -a(x). Since \overline{a}(x+y) = -a(x+y) = -a(x)-a(y) = \overline{a}(x) + \overline{a}(y), \overline{a} is a group homomorphism. Moreover, (a + \overline{a})(x) = a(x) + \overline{a}(x) = a(x) - a(x) = 0 = 0(x), so that a + \overline{a} = 0. Similarly, \overline{a} + a = 0. Thus every element has an additive inverse.
  4. Note that (a + b)(x) = a(x) + b(x) = b(x) + a(x) = (b+a)(x), so that a+b = b+a, and R is an abelian group under addition.
  5. Recall that function composition is always associative, so that multiplication is associative.
  6. Note that (a \circ (b+c))(x) = a((b+c)(x)) a(b(x) + c(x)) = a(b(x)) + a(c(x)) = (a \circ b)(x) + (a \circ c)(x) = (a \circ b + a \circ c)(x). Thus a \circ (b + c) = a \circ b + a \circ c, and multiplication distributes over addition on the left.
  7. Similarly, ((a + b) \circ c)(x) = (a+b)(c(x)) = a(c(x)) + b(c(x)) = (a \circ c)(x) + (b \circ c)(x) = (a \circ c + b \circ c)(x), and we have (a+b) \circ c = a \circ c + b \circ c. Thus multiplication distributes over addition on the right.
  8. Finally, let 1 denote the identity homomorphism. Clearly 1 \cdot a = a = a \cdot 1, so that 1 is a multiplicative identity element.
  9. If 1 = 0, then |A| = 1. If |A| = 1, then clearly 1 = 0. That is, 1 \neq 0 if and only if A is nontrivial.

Thus R is a ring with 1, and 1 \neq 0 if and only if A is nontrivial.

Now we consider the units in R.

If a \in R is a unit, then there exists a homomorphism b : A \rightarrow A such that a \circ b = b \circ a = 1. Thus a is injective and surjective, and in fact a is an automorphism of A. Conversely, if a is an automorphism of A, then a^{-1} is an automorphism, and we have a \circ a^{-1} = a^{-1} \circ a = 1. Thus a \in R is a unit. That is, the units of R are precisely the group automorphisms of A.

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