hom set from R to a Group as R-Module

1. Proposition

Let \(R\) be a ring, \(A\) a group and

\begin{align*} \mathrm{Hom}_{\mathrm{Grp}}(R, A) \end{align*}

the set of group-homomorphisms. Then \(\mathrm{Hom}_{\mathrm{Grp}}(R, A)\) is a \(R\)-module given by

\begin{align*} r \cdot f(-) = f(r \cdot -) \end{align*}

2. Proof

2.1. welldefined group homomorphism

Suppose \(r_1,r_2 \in R\), then

\begin{align*} r \cdot f(r_1 + r_2) =& f(r \cdot (r_1 + r_2)) \\ =& f(r \cdot r_1 + r \cdot r_2) \\ =& f(r \cdot r_1) + f(r \cdot r_2) \\ =& r \cdot f(r_1) + r \cdot f(r_2) \end{align*}

2.2. r-module

Follows from

\begin{align*} (r_2 \cdot r_1) \cdot f(-) =& f(r_2 \cdot r_1 \cdot -) \\ =& f(r_2 \cdot (r_1 \cdot -)) \\ =& r_2 \cdot f(r_1 \cdot -) \\ =& r_2 \cdot (r_1 \cdot f(-)) \end{align*}

and

\begin{align*} 1 \cdot f(-) =& f(1 \cdot -) \\ =& f(-) \end{align*}