## 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 $A$.