On group theory 7

[Papapetros Vaggelis]

Let \(\displaystyle{\left(G,+\right)}\) be an abelian group.

Find the cardinality of \(\displaystyle{\rm{Hom}(\mathbb{Z},G)}\).

[Demetres]

Isn't it easy that the cardinality is $|G|$ regardless of whether $G$ is abelian or not?

[Papapetros Vaggelis]

Hello mr.Demetres.

Here is a more analytical solution.

Not only, \(\displaystyle{\rm{card}(\rm{Hom}(\mathbb{Z},G))=|G|}\), but

\(\displaystyle{\left(G,+\right)\simeq \left(\rm{Hom}(\mathbb{Z},G),+\right)}\) as abelian groups.

Indeed, let \(\displaystyle{f\in\rm{Hom}(\mathbb{Z},G)}\) . Then,

\(\displaystyle{\forall\,n\in\mathbb{Z}: f(n)=f(n\cdot 1)=n\cdot f(1)}\) and \(\displaystyle{f(1)\in G}\) .

On the other hand, if \(\displaystyle{f:\mathbb{Z}\longrightarrow G}\) is a function defined by

\(\displaystyle{f(n)=n\,g}\) for some \(\displaystyle{g\in G}\), then

\(\displaystyle{f(n+m)=(n+m)\,g=n\,g+m\,g=f(n)+f(m)\,,\forall\,n\,,m\in\mathbb{Z}}\), which means

that \(\displaystyle{f\in\rm{Hom}(\mathbb{Z},G)}\) .

We define \(\displaystyle{\Phi:\left(G,+\right)\longrightarrow \left(\rm{Hom}(\mathbb{Z},G),+\right)}\)

by \(\displaystyle{\Phi(g):\mathbb{Z}\longrightarrow G\,,\Phi(g)(n)=n\,g\,,\forall\,g\in G}\) .

According to the above analysis, the function \(\displaystyle{\Phi}\) is well defined and onto \(\displaystyle{\rm{Hom}(\mathbb{Z},G)}\) .

If \(\displaystyle{g_1\,,g_2\in G}\), then :

\(\displaystyle{\begin{aligned} \Phi(g_1+g_2)(n)&=n\,(g_1+g_2)\\&=n\,g_1+n\,g_2\\&=\Phi(g_1)(n)+\Phi(g_2)(n)\\&=\left(\Phi(g_1)+\Phi(g_2)\right)(n)\,,\forall\,n\in\mathbb{Z}\end{aligned}}\)

so, \(\displaystyle{\Phi(g_1+g_2)=\Phi(g_1)+\Phi(g_2)\,,\forall\,g_1\,,g_2\in G}\), which means that

the function \(\displaystyle{\Phi}\) is homomorphism.

Finally,

\(\displaystyle{\begin{aligned} \rm{Ker}(\Phi)&=\left\{g\in G: \Phi(g)=\mathbb{O}\right\}\\&=\left\{g\in G: \Phi(g)(n)=0_{G}\,,\forall\,n\in\mathbb{Z}\right\}\\&=\left\{g\in G: n\,g=0\,,\forall\,n\in\mathbb{Z}\right\}\\&=\left\{0_{G}\right\}\end{aligned}}\)

Therefore,

\(\displaystyle{\left(G,+\right)\simeq \left(\rm{Hom}(\mathbb{Z},G),+\right)}\) .

[Papapetros Vaggelis]

In general, \(\displaystyle{\rm{card}(\rm{Hom}(\mathbb{Z}^{n},G))=\rm{card}(G^{n})\,\,,\forall\,n\in\mathbb{N}}\) .