3.3 The set of homomorphisms
From now on, when we write \(\mathbb {Z}\) in the context when a group is expected, we always mean the group \((\mathbb {Z},+)\). In other words, the binary operation is always assumed to be addition unless we explicitly say otherwise.
Suppose \(G\) and \(H\) are groups. We write \(\operatorname{\mathrm{Hom}}(G,H)\) for the set of all group homomorphisms \(f:G\to H\).
Suppose \(f_1,f_2\in \operatorname{\mathrm{Hom}}(G,H)\) for groups \(G\) and \(H\). Set \(K = \{ g\in G: f_1(g) = f_2(g)\} \). Then \(K\leq G\).
Write \(e\) for the idenity elements of \(G\) and \(H\) (by abuse of notation). Since \(f_1,f_2\in \operatorname{\mathrm{Hom}}(G,H)\), \(f_1(e) = e = f_2(e)\). So \(e\in K\). Thus \(K\neq \emptyset \) and we can use the one-step subgroup test. So suppose \(x,y\in K\). Then \(f_1(xy^{-1}) = f_1(x)f_1(y)^{-1} = f_2(x)f_2(y)^{-1} = f_2(xy^{-1})\). So \(xy^{-1}\in K\). So \(K\leq G\).
For arbitary \(G\) and \(H\), it is usually rather difficult to determine what exactly \(\operatorname{\mathrm{Hom}}(G,H)\) is. However, we do have the following wonderful (but easy) theorem.
Suppose \(G\) is a group. Then the map
is a bijection. In particular, for every \(g\in G\), there exists a unique group homomorphism \(f:\mathbb {Z}\to G\) such that \(f(1) = g\). In fact, this homomorphism is given by \(f(n) = g^n\).
To prove that \(\epsilon \) is one-one, suppose \(\epsilon (f_1) = \epsilon (f_2)\), and set \(K = \{ n\in N: f_1(n) = f_2(n)\} \). By Lemma 3.15, \(K\leq \mathbb {Z}\). But, by assumption, \(1\in K\). So \(\mathbb {Z} = \langle 1\rangle \leq K\leq \mathbb {Z}\), and, therefore, \(K = \mathbb {Z}\). So \(f_1 = f_2\).
To show that, \(\epsilon \) is onto suppose \(g\in G\). Define a map \(f:\mathbb {Z}\to G\) given by \(f(n) = g^n\). Then, for \(n,m\in \mathbb {Z}\), \(f(n+m) = g^{n+m} = g^n g^m = f(n)f(m)\). So \(f\in \operatorname{\mathrm{Hom}}(\mathbb {Z},G)\), and \(\epsilon (f) = f(1) = g\).