Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.29
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 $A$.
Solution: Let $a,b,c \in R$, and let $x \in A$.
(1) Note that \begin{align*}(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).\end{align*} 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 \begin{align*}(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).\end{align*} Thus $a \circ (b + c) = a \circ b + a \circ c$, and multiplication distributes over addition on the left.
(7) Similarly, \begin{align*}((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),\end{align*} 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 $A$.
