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Suppose $G$ abelian and $f:G\rightarrow \mathbb Z$ is surjective with kernel $K$.

  • Show that $G$ has a subgroup $H$ such that $H \cong \mathbb{Z}$
  • Show that $G \cong H\bigoplus K$

To get started:

There exists $g \epsilon G$ such that $f(g)=1$. Now set $H=<g>$. Clearly $H \cong \Bbb Z$.

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This is false. Consider $\mathbb{Z}/p^2 = G$ and $f$ multiplication by $p.$ Then $G \not\cong ker(f) \oplus im(f) \cong \mathbb{Z}/p \oplus \mathbb{Z}/p.$ –  jspecter Oct 31 '12 at 2:20
    
$Z$ was changed to $\Bbb Z$...should work...formatting lapse on my part. –  KUSH Oct 31 '12 at 3:00
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3 Answers

up vote 3 down vote accepted

If you knew anything about projective modules, this result is pretty much trivial.

If you don't, I think it's still not bad. Let $h\in G$ be arbitrary (but fixed) such that $f(h)=1$ (such an $h$ exists because $f$ is onto). It is easy to show $|h|$ (the order of $h$) is infinite (otherwise, $0=f(0)=f(|h|h)=|h|f(h)=|h|\cdot 1=|h|>0$). Thus, $G\geq\langle h\rangle\cong\mathbb{Z}$. Denote $\langle h\rangle$ by $H$.

If, $g\in G$ is arbitrtary, say $f(g)=n\in\mathbb{Z}$. Write $g=nh+g-nh=nh+(g-nh)$. Obviously $nh\in H$, and $f(g-nh)=f(g)-nf(h)=n-n\cdot 1=n-n=0$, so $g-nh\in \mathrm{ker}f=K$. This shows $G=H+K$. To show this is a direct sum, it suffices to show $H\cap K=\{0\}$. This is clear because no multiple of $h$ may map to $0$ (by the same argument from the previous paragraph).

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Thanks...I don't know projective modules but that makes sense! –  KUSH Oct 31 '12 at 3:14
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Since $f$ is surjective, there exists $a \in G$ such that $f(a) = 1$. Let $g\colon \mathbb{Z} \rightarrow G$ be a unique homomorphism such that $g(1) = a$. Since $fg = 1$, $g$ is injective. Let $H = g(\mathbb{Z})$. Then $H$ is isomorphic to $\mathbb{Z}$. For $x \in G$, $f(x - gf(x)) = f(x) - f(x) = 0$. Hence $x - gf(x) \in K$. Hence $G = H + K$. Suppose $y \in H \cap K$. There exists $n \in \mathbb{Z}$ such that $y = g(n)$. Since $0 = f(y) = fg(n) = n$, $y = 0$. Hence $G = H \oplus K$.

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Thanks, very clear! –  KUSH Oct 31 '12 at 3:15
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Let us look at the short exact sequence

$$0\rightarrow \ker f\stackrel{i}\rightarrow G\stackrel{f}\rightarrow \Bbb Z\rightarrow 0$$

with $\,i\,$ injection.

Let us use the universal property of the free abelian group $\,\Bbb Z=\langle\,1\,\rangle\,$ : choose $\,x\in G\setminus\ker f\,$ and put $\,\overline g(1):=x\,$ , then there exists a unique homomorphism $\,g:\Bbb Z\to G\,$ s.t. $\,g(1)=\overline g(1)=x\,$.

Since $\,\overline 0\neq x+\ker f\in G/\ker f\cong\Bbb Z\,$ , clearly $\,x\,$ has infinite order and thus

$$H:=g(\Bbb Z)=\langle\,x\,\rangle\,\cong\Bbb Z$$

Now : $$y\in H\cap \ker f\Longrightarrow y=mx\wedge f(y)=0\;\;,\;m\in\Bbb Z\Longrightarrow$$

$$ 0=f(mx)=mf(x)\Longrightarrow m=0\vee x\in \ker f$$

Since $\,x\notin\ker f\,$ by choice, it must be $\,m=0\Longrightarrow y=0\,$

Can you take it from here now and end the argument?

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Since the full answer has been accepted, here's a great exposition of the end of the argument: math.uconn.edu/~kconrad/blurbs/grouptheory/splittinggp.pdf –  Alexei Averchenko Oct 31 '12 at 5:19
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