Question

Question: If G is an abelian group and $\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathbf{\mathbb{Z}}$a surjective homomorphism with kernel K, prove that G has a subgroup H such that$\mathit{H}\mathbf{\cong }\mathbf{\mathbb{Z}}$and $\mathit{G}\mathbf{=}\mathit{H}\mathbf{\oplus }\mathit{K}$.

Short answer

It is proved that and .

Step-by-step solution

Theorem 9.3

If M and N are normal subgroups of a group $\mathit{G}\mathbf{=}\mathit{M}\mathit{N}$and $\mathit{M}\mathbf{\cap }\mathit{N}\mathbf{=}<e>$, then$\mathit{G}\mathbf{=}\mathit{M}\mathbf{\times }\mathit{N}$.

Given that G is an abelian group and$f:G\to \mathrm{\mathbb{Z}}$ a surjective homomorphism with kernel K.

Prove that G=H⊕K

Let$g_1\in G$be any element such that,$f\left(g_1\right)=1$.

Define.

We will prove that $G=H+K$and.

Let $g\in G$.

It is easy to see that, $g=\left(g-f\left(g\right)g_1\right)+f\left(g\right)g$.

Note that,$f\left(g\right)g_1\in H$and find $f\left(g-f\left(g\right)g_1\right)$as:

$$\begin{gathered} f(g-f(g\left)g_1\right)=f\left(g\right)-f\left(g\right)f\left(g_1\right) \\ =f\left(g\right)-f\left(g\right).1 \\ =f\left(g\right)-f\left(g\right) \\ =0 \end{gathered}$$

So that$g-f\left(g\right)g_1\in K$.

Since$g$is chosen arbitrarily,$G=H+K$.

Now, let$g\in H\cap K$.

Then, there is some$n\in \mathrm{\mathbb{Z}}$such thatlocalid="1656589739620" $g=n{g}_1$and$f\left(g\right)=0$.

Simplify$f\left(g\right)=0$as:

$$\begin{gathered} 0=f\left(g\right) \\ =f\left(n{g}_1\right) \\ =nf\left(g_1\right) \\ =n.1 \end{gathered}$$

Therefore,$n=0$, which implies$g=o{g}_1=0$$g=0g_1=0$.

We conclude that, .

Hence, by theoremTheorem 9.3,$G=H\oplus K$ .